CLASS 12 MATHS · CHAPTER 5
Continuity and Differentiabilityसांतत्य तथा अवकलनीयताContinuity aur Differentiability
Continuity (graphs without breaks), differentiability, chain rule, implicit and inverse-trig derivatives, exponential & log functions, logarithmic differentiation, parametric forms and second order derivatives — NCERT Chapter 5, in easy words.सांतत्य (बिना टूटे ग्राफ़), अवकलनीयता, श्रृंखला नियम, अस्पष्ट और प्रतिलोम त्रिकोणमितीय फलनों के अवकलज, चरघातांकी तथा लघुगणकीय फलन, लघुगणकीय अवकलन, प्राचलिक रूप और द्वितीय कोटि के अवकलज — NCERT अध्याय 5, आसान शब्दों में।Continuity (bina toote graph), differentiability, chain rule, implicit aur inverse-trig derivatives, exponential aur log functions, logarithmic differentiation, parametric form aur second order derivatives — NCERT Chapter 5, aasaan shabdon mein.
In this chapterइस अध्याय मेंIs chapter mein
Didn’t understand a paragraph? Tap its अ button → हिंदी / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उसके अ बटन पर टैप कीजिए → English / Hinglish / Simple English।Koi paragraph samajh nahi aaya? Uske अ button par tap kijiye → हिंदी / English / Simple English.
Pictures — tap to zoomचित्र — टैप करके बड़ा कीजिएPictures — tap karke bada kijiye
A jump: LHL ≠ RHL at x = 0छलाँग: x = 0 पर LHL ≠ RHLJump: x = 0 par LHL ≠ RHLFig 5.1
Limits agree, but f(0) is somewhere elseसीमाएँ बराबर, पर f(0) कहीं औरLimits barabar, par f(0) kahin aurFig 5.2
y = 1/x near 0: shoots to +∞ and −∞0 के पास y = 1/x: +∞ और −∞ की ओर0 ke paas y = 1/x: +∞ aur −∞ ki tarafFig 5.3
Example 10: jump at x = 1उदाहरण 10: x = 1 पर छलाँगExample 10: x = 1 par jumpFig 5.4
Greatest integer function [x]: a jump at every integerमहत्तम पूर्णांक फलन [x]: हर पूर्णांक पर छलाँगGreatest integer function [x]: har integer par jumpFig 5.8
Secant → tangent: the derivative is a limit of slopesछेदक रेखा → स्पर्श रेखा: अवकलज ढालों की सीमा हैSecant → tangent: derivative slopes ki limit haiSection 5.3
f(x) = |x|: continuous at 0 but not differentiable theref(x) = |x|: 0 पर संतत पर अवकलनीय नहींf(x) = |x|: 0 par continuous par differentiable nahiSection 5.3
Chain rule: a box inside a boxश्रृंखला नियम: डिब्बे के अंदर डिब्बाChain rule: dabbe ke andar dabbaTheorem 4
y = eˣ and y = log x: mirror images in y = xy = eˣ और y = log x: y = x में दर्पण प्रतिबिंबy = eˣ aur y = log x: y = x mein mirror imageFig 5.11
A parametric curve: x = a(θ + sin θ), y = a(1 − cos θ)प्राचलिक वक्र: x = a(θ + sin θ), y = a(1 − cos θ)Parametric curve: x = a(θ + sin θ), y = a(1 − cos θ)Example 33Try it yourselfखुद करके देखिएKhud karke dekhiye
All sections (NCERT order)सभी खंड (NCERT क्रम)Saare sections (NCERT order)
- 5.1IntroductionभूमिकाIntroduction
- 5.2Continuityसांतत्यContinuity
- 5.2.1Algebra of continuous functionsसंतत फलनों का बीजगणितContinuous functions ka algebra
- 5.3DifferentiabilityअवकलनीयताDifferentiability
- 5.3.1Derivatives of composite functionsसंयुक्त फलनों के अवकलजComposite functions ke derivatives
- 5.3.2Derivatives of implicit functionsअस्पष्ट फलनों के अवकलजImplicit functions ke derivatives
- 5.3.3Derivatives of inverse trigonometric functionsप्रतिलोम त्रिकोणमितीय फलनों के अवकलजInverse trigonometric functions ke derivatives
- 5.4Exponential and Logarithmic Functionsचरघातांकी तथा लघुगणकीय फलनExponential aur Logarithmic Functions
- 5.5Logarithmic Differentiationलघुगणकीय अवकलनLogarithmic Differentiation
- 5.6Derivatives of Functions in Parametric Formsप्राचलिक रूपों में फलनों के अवकलजParametric form mein functions ke derivatives
- 5.7Second Order Derivativeद्वितीय कोटि का अवकलजSecond Order Derivative
- Miscellaneous Examplesविविध उदाहरणMiscellaneous Examples
- Formula sheetसूत्र-पत्रFormula sheet
- Quick revisionत्वरित पुनरावृत्तिQuick revision
5.1IntroductionभूमिकाIntroduction
This chapter continues the study of differentiation of functions which you began in Class XI, where you differentiated polynomial and trigonometric functions. Here we introduce the very important ideas of continuity, differentiability and the relation between them. We also learn to differentiate inverse trigonometric functions, and a new class of functions called exponential and logarithmic functions.यह अध्याय अवकलन के उस अध्ययन को आगे बढ़ाता है जो आपने कक्षा XI में बहुपद और त्रिकोणमितीय फलनों का अवकलन करके शुरू किया था। यहाँ हम सांतत्य, अवकलनीयता और इनके बीच के संबंध की महत्वपूर्ण अवधारणाएँ सीखेंगे। साथ ही प्रतिलोम त्रिकोणमितीय फलनों का अवकलन, और चरघातांकी तथा लघुगणकीय फलन नामक नए फलनों का अवकलन भी सीखेंगे।Ye chapter differentiation ki wahi padhai aage badhata hai jo aapne Class XI mein polynomial aur trigonometric functions ko differentiate karke shuru ki thi. Yahan hum continuity, differentiability aur inke beech ka relation seekhenge. Saath hi inverse trigonometric functions ka derivative, aur exponential aur logarithmic functions naam ke naye functions ka derivative bhi seekhenge.We learn continuity, differentiability, and derivatives of inverse-trig, exponential and log functions.
5.2Continuityसांतत्यContinuity
Two informal examples first. Let f(x) = 1 if x \le 0 and f(x) = 2 if x > 0 (Fig 5.1). Near x = 0, at points to the left like -0.1, -0.01, -0.001 the value is 1; at points to the right like 0.1, 0.01, 0.001 the value is 2. So the left hand limit at 0 is 1 and the right hand limit is 2 — they do not coincide. The graph cannot be drawn in one stroke: we must lift the pen at 0. This function is not continuous at x = 0.पहले दो अनौपचारिक उदाहरण। मान लीजिए x \le 0 के लिए f(x) = 1 और x > 0 के लिए f(x) = 2 (आकृति 5.1)। x = 0 के पास, बाईं ओर के बिंदुओं -0.1, -0.01, -0.001 पर मान 1 है; दाईं ओर के बिंदुओं 0.1, 0.01, 0.001 पर मान 2 है। अतः 0 पर बाएँ पक्ष की सीमा 1 और दाएँ पक्ष की सीमा 2 है — दोनों बराबर नहीं। ग्राफ़ एक बार में नहीं खींचा जा सकता: 0 पर पेन उठाना पड़ता है। यह फलन x = 0 पर संतत नहीं है।Pehle do informal examples. Maan lijiye x \le 0 ke liye f(x) = 1 aur x > 0 ke liye f(x) = 2 (Fig 5.1). x = 0 ke paas, left ke points -0.1, -0.01, -0.001 par value 1 hai; right ke points 0.1, 0.01, 0.001 par value 2 hai. To 0 par LHL 1 aur RHL 2 hai — dono barabar nahi. Graph ek stroke mein nahi banta: 0 par pen uthana padta hai. Ye function x = 0 par continuous nahi hai.A jump: LHL = 1, RHL = 2, so not continuous at 0.
A jump: LHL ≠ RHL at x = 0छलाँग: x = 0 पर LHL ≠ RHLJump: x = 0 par LHL ≠ RHLNCERT Fig 5.1 · 8 tappable partsOpenNow let f(x) = 1 if x \ne 0 and f(x) = 2 if x = 0 (Fig 5.2). Here the left and right hand limits at 0 are both 1, but the value f(0) = 2 does not coincide with them. Again we cannot draw the graph without lifting the pen — another function not continuous at x = 0.अब मान लीजिए x \ne 0 के लिए f(x) = 1 और x = 0 के लिए f(x) = 2 (आकृति 5.2)। यहाँ 0 पर बाएँ और दाएँ पक्ष की सीमाएँ दोनों 1 हैं, पर मान f(0) = 2 उनसे मेल नहीं खाता। फिर से पेन उठाए बिना ग्राफ़ नहीं बनता — यह फलन भी x = 0 पर संतत नहीं है।Ab maan lijiye x \ne 0 ke liye f(x) = 1 aur x = 0 ke liye f(x) = 2 (Fig 5.2). Yahan 0 par LHL aur RHL dono 1 hain, par value f(0) = 2 unse match nahi karti. Phir se pen uthaye bina graph nahi banta — ye function bhi x = 0 par continuous nahi hai.A moved dot: LHL = RHL = 1 but f(0) = 2, so not continuous.
Limits agree, but f(0) is somewhere elseसीमाएँ बराबर, पर f(0) कहीं औरLimits barabar, par f(0) kahin aurNCERT Fig 5.2 · 6 tappable partsOpenSolution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- First check that f is defined at the point: yes, its value is 5.पहले देखिए कि f उस बिंदु पर परिभाषित है: हाँ, मान 5 है।Pehle dekhiye ki f us point par defined hai: haan, value 5 hai.f(1) = 2(1) + 3 = 5
- Find the limit at x = 1.x = 1 पर सीमा निकालिए।x = 1 par limit nikaaliye.\lim_{x \to 1} f(x) = \lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5
- Limit equals the value.सीमा = मान।Limit = value.\lim_{x \to 1} f(x) = 5 = f(1)
✔ f is continuous at x = 1.f, x = 1 पर संतत है।f, x = 1 par continuous hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- f is defined at 0.f, 0 पर परिभाषित है।f, 0 par defined hai.f(0) = 0^2 = 0
- Limit equals the value.सीमा = मान।Limit = value.\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 = 0^2 = 0 = f(0)
✔ f is continuous at x = 0.f, x = 0 पर संतत है।f, x = 0 par continuous hai.
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Write |x| in two pieces (definition of modulus).|x| को दो टुकड़ों में लिखिए (मापांक की परिभाषा)।|x| ko do tukdon mein likhiye (modulus ki definition).f(x) = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}
- Value at 0.0 पर मान।0 par value.f(0) = 0
- Left of 0 use -x, right of 0 use x.0 के बाईं ओर -x, दाईं ओर x लीजिए।0 ke left -x, right x lijiye.\text{LHL} = \lim_{x \to 0^-} (-x) = 0, \quad \text{RHL} = \lim_{x \to 0^+} x = 0
- All three coincide.तीनों बराबर हैं।Teeno barabar hain.\text{LHL} = \text{RHL} = f(0) = 0
✔ f is continuous at x = 0.f, x = 0 पर संतत है।f, x = 0 par continuous hai.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- The value at 0 is given separately.0 पर मान अलग से दिया है।0 par value alag se di hai.f(0) = 1
- Near 0 (but not at 0) f is the polynomial x^3 + 3.0 के पास (पर 0 पर नहीं) f बहुपद x^3 + 3 है।0 ke paas (par 0 par nahi) f polynomial x^3 + 3 hai.\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^3 + 3) = 0^3 + 3 = 3
- Limit does not coincide with the value.सीमा, मान के बराबर नहीं।Limit, value ke barabar nahi.\lim_{x \to 0} f(x) = 3 \ne 1 = f(0)
✔ f is not continuous at x = 0; x = 0 is its only point of discontinuity.f, x = 0 पर संतत नहीं है; x = 0 इसका एकमात्र असांतत्य का बिंदु है।f, x = 0 par continuous nahi hai; x = 0 iska akela point of discontinuity hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Defined at all real numbers, always equal to k.सभी वास्तविक संख्याओं पर परिभाषित, मान हमेशा k।Sabhi real numbers par defined, value hamesha k.f(c) = k for every real c
- Limit of a constant is the constant.अचर की सीमा वही अचर है।Constant ki limit wahi constant hai.\lim_{x \to c} f(x) = \lim_{x \to c} k = k = f(c)
✔ The constant function is continuous at every real number.अचर फलन प्रत्येक वास्तविक संख्या पर संतत है।Constant function har real number par continuous hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Defined everywhere.हर जगह परिभाषित।Har jagah defined.f(c) = c
- Limit equals the value at every c.हर c पर सीमा = मान।Har c par limit = value.\lim_{x \to c} f(x) = \lim_{x \to c} x = c = f(c)
✔ The identity function is continuous at every real number.तत्समक फलन प्रत्येक वास्तविक संख्या पर संतत है।Identity function har real number par continuous hai.
If f is defined on a closed interval [a, b], then f must be continuous at every point of [a, b] including the end points. At the end points only one side makes sense: continuity at a means \lim_{x \to a^+} f(x) = f(a) and continuity at b means \lim_{x \to b^-} f(x) = f(b). (\lim_{x \to a^-} f(x) and \lim_{x \to b^+} f(x) do not make sense.) As a consequence, if the domain of f is a single point, f is continuous.यदि f बंद अंतराल [a, b] पर परिभाषित है, तो f को [a, b] के हर बिंदु पर, अंत्य बिंदुओं सहित, संतत होना चाहिए। अंत्य बिंदुओं पर केवल एक ओर का अर्थ बनता है: a पर सांतत्य का अर्थ \lim_{x \to a^+} f(x) = f(a) और b पर सांतत्य का अर्थ \lim_{x \to b^-} f(x) = f(b)। (\lim_{x \to a^-} f(x) और \lim_{x \to b^+} f(x) का कोई अर्थ नहीं।) परिणामस्वरूप, यदि f का प्रांत एकल बिंदु है, तो f संतत है।Agar f closed interval [a, b] par defined hai, to f ko [a, b] ke har point par, end points samet, continuous hona chahiye. End points par sirf ek side ka matlab banta hai: a par continuity ka matlab \lim_{x \to a^+} f(x) = f(a) aur b par \lim_{x \to b^-} f(x) = f(b). (\lim_{x \to a^-} f(x) aur \lim_{x \to b^+} f(x) ka koi matlab nahi.) Isliye, agar f ka domain ek hi point hai, to f continuous hai.On [a, b], check the right limit at a and the left limit at b.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Rewrite in pieces; by Example 3, f is continuous at x = 0.टुकड़ों में लिखिए; उदाहरण 3 से f, x = 0 पर संतत है।Tukdon mein likhiye; Example 3 se f, x = 0 par continuous hai.f(x) = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}
- Near a negative c, all nearby x are negative too, so f is -x there.किसी ऋणात्मक c के पास सभी x भी ऋणात्मक हैं, इसलिए वहाँ f = -x।Kisi negative c ke paas saare x bhi negative hain, isliye wahan f = -x.c < 0:\ \lim_{x \to c} (-x) = -c = f(c)
- Near a positive c, f is x.किसी धनात्मक c के पास f = x।Kisi positive c ke paas f = x.c > 0:\ \lim_{x \to c} x = c = f(c)
✔ Yes. f is continuous at all points, so |x| is a continuous function.हाँ। f सभी बिंदुओं पर संतत है, अतः |x| एक संतत फलन है।Haan. f saare points par continuous hai, isliye |x| ek continuous function hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Defined at every real c.हर वास्तविक c पर परिभाषित।Har real c par defined.f(c) = c^3 + c^2 - 1
- Limit of a polynomial: just substitute.बहुपद की सीमा: सीधे मान रखिए।Polynomial ki limit: seedha value rakhiye.\lim_{x \to c} (x^3 + x^2 - 1) = c^3 + c^2 - 1 = f(c)
✔ f is continuous at every real number, i.e., f is a continuous function.f प्रत्येक वास्तविक संख्या पर संतत है, अर्थात् f एक संतत फलन है।f har real number par continuous hai, yaani f ek continuous function hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- The domain is all non-zero reals; take any point c of it.प्रांत सभी शून्येतर वास्तविक संख्याएँ हैं; उसका कोई बिंदु c लीजिए।Domain saare non-zero real numbers hain; uska koi point c lijiye.\text{Fix any } c \ne 0:\ \lim_{x \to c} \dfrac{1}{x} = \dfrac{1}{c}
- Limit equals value at every point of the domain.प्रांत के हर बिंदु पर सीमा = मान।Domain ke har point par limit = value.f(c) = \dfrac{1}{c} = \lim_{x \to c} f(x)
✔ f is continuous at every point of its domain, so f(x) = \dfrac{1}{x} is a continuous function. (x = 0 is not in its domain, so it is not a "point of discontinuity" by this definition.)f अपने प्रांत के हर बिंदु पर संतत है, अतः f(x) = \dfrac{1}{x} एक संतत फलन है। (x = 0 इसके प्रांत में ही नहीं है।)f apne domain ke har point par continuous hai, isliye f(x) = \dfrac{1}{x} ek continuous function hai. (x = 0 iske domain mein hi nahi hai.)
What does infinity mean here? Look at f(x) = \dfrac{1}{x} near 0. At x = 1, 0.3, 0.2, 0.1, 0.01, 0.001, 10^{-n} the values are 1, 3.333\ldots, 5, 10, 100, 1000, 10^{n} (Table 5.1): as x comes close to 0 from the right, f(x) shoots up beyond any number. We write \lim_{x \to 0^+} f(x) = +\infty. From the left (Table 5.2: x = -1, -0.3, \ldots, -10^{-n} give -1, -3.333\ldots, \ldots, -10^{n}) we write \lim_{x \to 0^-} f(x) = -\infty. Remember: +\infty and -\infty are NOT real numbers, so these one-sided limits do not exist as real numbers (Fig 5.3).यहाँ अनंत का क्या अर्थ है? 0 के पास f(x) = \dfrac{1}{x} देखिए। x = 1, 0.3, 0.2, 0.1, 0.01, 0.001, 10^{-n} पर मान 1, 3.333\ldots, 5, 10, 100, 1000, 10^{n} हैं (सारणी 5.1): जैसे-जैसे x दाईं ओर से 0 के पास आता है, f(x) किसी भी संख्या से बड़ा हो जाता है। हम लिखते हैं \lim_{x \to 0^+} f(x) = +\infty। बाईं ओर से (सारणी 5.2) \lim_{x \to 0^-} f(x) = -\infty। याद रखिए: +\infty और -\infty वास्तविक संख्याएँ नहीं हैं, इसलिए ये एकपक्षीय सीमाएँ वास्तविक संख्या के रूप में विद्यमान नहीं हैं (आकृति 5.3)।Yahan infinity ka kya matlab hai? 0 ke paas f(x) = \dfrac{1}{x} dekhiye. x = 1, 0.3, 0.2, 0.1, 0.01, 0.001, 10^{-n} par values 1, 3.333\ldots, 5, 10, 100, 1000, 10^{n} hain (Table 5.1): jaise-jaise x right se 0 ke paas aata hai, f(x) kisi bhi number se bada ho jaata hai. Hum likhte hain \lim_{x \to 0^+} f(x) = +\infty. Left se (Table 5.2) \lim_{x \to 0^-} f(x) = -\infty. Yaad rakhiye: +\infty aur -\infty real numbers nahi hain, isliye ye one-sided limits real number ke roop mein exist nahi karti (Fig 5.3).Near 0, 1/x grows without bound; ±∞ are not real numbers, so these limits do not exist.
y = 1/x near 0: shoots to +∞ and −∞0 के पास y = 1/x: +∞ और −∞ की ओर0 ke paas y = 1/x: +∞ aur −∞ ki tarafNCERT Fig 5.3 · 7 tappable partsOpenSolution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Case 1: left of 1, f is the polynomial x + 2 — continuous.स्थिति 1: 1 के बाईं ओर f बहुपद x + 2 है — संतत।Case 1: 1 ke left f polynomial x + 2 hai — continuous.c < 1:\ \lim_{x \to c} (x + 2) = c + 2 = f(c)
- Case 2: right of 1, f is x - 2 — continuous.स्थिति 2: 1 के दाईं ओर f = x - 2 — संतत।Case 2: 1 ke right f = x - 2 — continuous.c > 1:\ \lim_{x \to c} (x - 2) = c - 2 = f(c)
- Case 3: at c = 1 the two pieces meet.स्थिति 3: c = 1 पर दोनों टुकड़े मिलते हैं।Case 3: c = 1 par dono tukde milte hain.\text{LHL} = \lim_{x \to 1^-} (x + 2) = 3, \quad \text{RHL} = \lim_{x \to 1^+} (x - 2) = -1
- Left and right hand limits do not coincide (a jump of 4).बाएँ और दाएँ पक्ष की सीमाएँ बराबर नहीं (4 की छलाँग)।LHL aur RHL barabar nahi (4 ki chhalaang).3 \ne -1
✔ f is not continuous at x = 1; x = 1 is the only point of discontinuity (Fig 5.4).f, x = 1 पर संतत नहीं है; x = 1 एकमात्र असांतत्य का बिंदु है (आकृति 5.4)।f, x = 1 par continuous nahi hai; x = 1 akela point of discontinuity hai (Fig 5.4).
Example 10: jump at x = 1उदाहरण 10: x = 1 पर छलाँगExample 10: x = 1 par jumpNCERT Fig 5.4 · 8 tappable partsOpenSolution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- As in Example 10, each piece is a polynomial.उदाहरण 10 की तरह, हर टुकड़ा बहुपद है।Example 10 ki tarah, har tukda polynomial hai.x \ne 1:\ \text{continuous}
- One-sided limits at 1.1 पर एकपक्षीय सीमाएँ।1 par one-sided limits.\text{LHL} = \lim_{x \to 1^-} (x + 2) = 3, \quad \text{RHL} = \lim_{x \to 1^+} (x - 2) = -1
- They do not coincide, so the value f(1) = 0 cannot save continuity.ये बराबर नहीं, इसलिए मान f(1) = 0 सांतत्य नहीं बचा सकता।Ye barabar nahi, isliye value f(1) = 0 continuity nahi bacha sakti.\text{LHL} \ne \text{RHL}
✔ x = 1 is the only point of discontinuity of f (Fig 5.5).x = 1, f का एकमात्र असांतत्य का बिंदु है (आकृति 5.5)।x = 1, f ka akela point of discontinuity hai (Fig 5.5).
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- f is not defined at 0, so 0 is not in the domain.f, 0 पर परिभाषित नहीं है, इसलिए 0 प्रांत में नहीं है।f, 0 par defined nahi hai, isliye 0 domain mein nahi hai.\text{Domain} = D_1 \cup D_2,\ D_1 = \{x < 0\},\ D_2 = \{x > 0\}
- Case 1: continuous in D_1.स्थिति 1: D_1 में संतत।Case 1: D_1 mein continuous.c \in D_1:\ \lim_{x \to c} (x + 2) = c + 2 = f(c)
- Case 2: continuous in D_2.स्थिति 2: D_2 में संतत।Case 2: D_2 mein continuous.c \in D_2:\ \lim_{x \to c} (-x + 2) = -c + 2 = f(c)
✔ f is continuous at every point of its domain, so f is a continuous function. We lift the pen only at x = 0, where f is not defined (Fig 5.6).f अपने प्रांत के हर बिंदु पर संतत है, अतः f एक संतत फलन है। पेन केवल x = 0 पर उठाना पड़ता है, जहाँ f परिभाषित ही नहीं है (आकृति 5.6)।f apne domain ke har point par continuous hai, isliye f ek continuous function hai. Pen sirf x = 0 par uthana padta hai, jahan f defined hi nahi hai (Fig 5.6).
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Split the real line into three disjoint parts.वास्तविक रेखा को तीन असंयुक्त भागों में बाँटिए।Real line ko teen alag hisson mein baantiye.D_1 = \{x < 0\},\ D_2 = \{0\},\ D_3 = \{x > 0\}
- Both are continuous there (Examples 2 and 6).दोनों वहाँ संतत हैं (उदाहरण 2 और 6)।Dono wahan continuous hain (Examples 2 aur 6).D_1:\ f(x) = x^2,\quad D_3:\ f(x) = x
- At 0, check all three numbers.0 पर तीनों संख्याएँ जाँचिए।0 par teeno numbers check kijiye.f(0) = 0,\ \lim_{x \to 0^-} x^2 = 0,\ \lim_{x \to 0^+} x = 0
✔ \lim_{x \to 0} f(x) = 0 = f(0), so f is continuous at 0 too; f is a continuous function (Fig 5.7).\lim_{x \to 0} f(x) = 0 = f(0), अतः f, 0 पर भी संतत है; f एक संतत फलन है (आकृति 5.7)।\lim_{x \to 0} f(x) = 0 = f(0), isliye f, 0 par bhi continuous hai; f ek continuous function hai (Fig 5.7).
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- A polynomial is defined for every real number.बहुपद प्रत्येक वास्तविक संख्या के लिए परिभाषित है।Polynomial har real number ke liye defined hai.p(x) = a_0 + a_1 x + \ldots + a_n x^n,\ a_n \ne 0,\ a_i \in \mathbf{R}
- For any fixed real c (limit of a polynomial = value).किसी भी नियत वास्तविक c के लिए (बहुपद की सीमा = मान)।Kisi bhi fixed real c ke liye (polynomial ki limit = value).\lim_{x \to c} p(x) = p(c)
✔ p is continuous at every real number, so every polynomial function is continuous.p प्रत्येक वास्तविक संख्या पर संतत है, अतः प्रत्येक बहुपद फलन संतत है।p har real number par continuous hai, isliye har polynomial function continuous hai.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Case 1: near a non-integer c, [x] stays equal to [c] — flat step, continuous.स्थिति 1: किसी अपूर्णांक c के पास [x] का मान [c] ही रहता है — समतल सीढ़ी, संतत।Case 1: kisi non-integer c ke paas [x] ki value [c] hi rehti hai — flat step, continuous.c \notin \mathbf{Z}:\ \lim_{x \to c} [x] = [c] = f(c)
- Case 2: for a small r > 0 just left of an integer the value is one less.स्थिति 2: छोटे r > 0 के लिए पूर्णांक के ठीक बाईं ओर मान एक कम है।Case 2: chhote r > 0 ke liye integer ke theek left value ek kam hai.c \in \mathbf{Z}:\ [c - r] = c - 1,\ [c + r] = c
- One-sided limits differ.एकपक्षीय सीमाएँ अलग हैं।One-sided limits alag hain.\lim_{x \to c^-} f(x) = c - 1 \ne c = \lim_{x \to c^+} f(x)
✔ The greatest integer function is discontinuous at every integer and continuous at every other real number (Fig 5.8).महत्तम पूर्णांक फलन प्रत्येक पूर्णांक पर असंतत है और बाकी हर वास्तविक संख्या पर संतत है (आकृति 5.8)।Greatest integer function har integer par discontinuous hai aur baaki har real number par continuous hai (Fig 5.8).
Greatest integer function [x]: a jump at every integerमहत्तम पूर्णांक फलन [x]: हर पूर्णांक पर छलाँगGreatest integer function [x]: har integer par jumpNCERT Fig 5.8 · 9 tappable partsOpen5.2.1Algebra of continuous functionsसंतत फलनों का बीजगणितContinuous functions ka algebra
Proof of (1). \lim_{x \to c} (f + g)(x) = \lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) = f(c) + g(c) = (f + g)(c) — using the definition of f + g, the theorem on limits, and continuity of f and g. The other parts are similar. Remarks: (i) taking f constant = \lambda in (3): \lambda g is continuous; with \lambda = -1, continuity of f gives continuity of -f. (ii) taking f constant in (4): \dfrac{\lambda}{g} is continuous wherever g(x) \ne 0; in particular \dfrac{1}{g} is.(1) की उपपत्ति। \lim_{x \to c} (f + g)(x) = \lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) = f(c) + g(c) = (f + g)(c) — f + g की परिभाषा, सीमाओं के प्रमेय और f, g के सांतत्य से। बाकी भाग इसी तरह हैं। टिप्पणी: (i) (3) में f को अचर = \lambda लेने पर \lambda g संतत है; \lambda = -1 से f संतत ⇒ -f संतत। (ii) (4) में f अचर लेने पर \dfrac{\lambda}{g} वहाँ संतत है जहाँ g(x) \ne 0; विशेष रूप से \dfrac{1}{g}।(1) ka proof. \lim_{x \to c} (f + g)(x) = \lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) = f(c) + g(c) = (f + g)(c) — f + g ki definition, limits ke theorem aur f, g ki continuity se. Baaki parts isi tarah hain. Remarks: (i) (3) mein f ko constant = \lambda lene par \lambda g continuous hai; \lambda = -1 se f continuous ⇒ -f continuous. (ii) (4) mein f constant lene par \dfrac{\lambda}{g} wahan continuous hai jahan g(x) \ne 0; khaas taur par \dfrac{1}{g}.Sum, difference, product and quotient (denominator not zero) of continuous functions are continuous.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- p and q are polynomials; the domain leaves out the zeros of q.p और q बहुपद हैं; प्रांत में q के शून्यक शामिल नहीं।p aur q polynomials hain; domain mein q ke zeros shaamil nahi.f(x) = \dfrac{p(x)}{q(x)},\ q(x) \ne 0
- By part (4) of Theorem 1.प्रमेय 1 के भाग (4) से।Theorem 1 ke part (4) se.p,\ q \text{ continuous (Example 14)} \Rightarrow \dfrac{p}{q} \text{ continuous}
✔ Every rational function is continuous (at every point of its domain).प्रत्येक परिमेय फलन संतत है (अपने प्रांत के हर बिंदु पर)।Har rational function continuous hai (apne domain ke har point par).
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Fact used (clear from the graph of sin x near 0; also \lim_{h \to 0} \cos h = 1).प्रयुक्त तथ्य (0 के पास sin x के ग्राफ़ से स्पष्ट; साथ ही \lim_{h \to 0} \cos h = 1)।Use kiya fact (0 ke paas sin x ke graph se saaf; saath hi \lim_{h \to 0} \cos h = 1).\lim_{x \to 0} \sin x = 0
- Put x = c + h; x \to c means h \to 0.x = c + h रखिए; x \to c का अर्थ h \to 0।x = c + h rakhiye; x \to c ka matlab h \to 0.x = c + h:\ \lim_{x \to c} \sin x = \lim_{h \to 0} \sin(c + h)
- Expand \sin(A + B).\sin(A + B) का प्रसार।\sin(A + B) ko expand kijiye.= \lim_{h \to 0} [\sin c \cos h + \cos c \sin h]
- Limits of each part.हर भाग की सीमा।Har part ki limit.= \sin c \cdot 1 + \cos c \cdot 0 = \sin c = f(c)
✔ \lim_{x \to c} \sin x = \sin c for every real c, so sine is a continuous function. (A similar proof works for cosine.)हर वास्तविक c के लिए \lim_{x \to c} \sin x = \sin c, अतः sine एक संतत फलन है। (cosine के लिए भी ऐसी ही उपपत्ति।)Har real c ke liye \lim_{x \to c} \sin x = \sin c, isliye sine ek continuous function hai. (Cosine ke liye bhi aisa hi proof.)
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Domain of tan.tan का प्रांत।Tan ka domain.\tan x = \dfrac{\sin x}{\cos x},\ \cos x \ne 0 \iff x \ne (2n + 1)\dfrac{\pi}{2}
- Quotient of continuous functions (Theorem 1(4)).संतत फलनों का भागफल (प्रमेय 1(4))।Continuous functions ka quotient (Theorem 1(4)).\sin,\ \cos \text{ continuous} \Rightarrow \dfrac{\sin}{\cos} \text{ continuous where defined}
✔ tan x is continuous at every point of its domain, i.e., it is a continuous function.tan x अपने प्रांत के हर बिंदु पर संतत है, अर्थात् यह एक संतत फलन है।tan x apne domain ke har point par continuous hai, yaani ye ek continuous function hai.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Break it as a composition (inner x^2, outer sin).इसे संयोजन के रूप में तोड़िए (भीतर x^2, बाहर sin)।Ise composition ki tarah todiye (andar x^2, bahar sin).f = g \circ h,\ g(x) = \sin x,\ h(x) = x^2
- Theorem 2.प्रमेय 2।Theorem 2.g,\ h \text{ continuous} \Rightarrow g \circ h \text{ continuous}
✔ \sin(x^2) is a continuous function.\sin(x^2) एक संतत फलन है।\sin(x^2) ek continuous function hai.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Name the inner and outer functions.भीतरी और बाहरी फलन के नाम रखिए।Andar aur bahar wale function ka naam rakhiye.g(x) = 1 - x + |x|,\ h(x) = |x|
- So f = h \circ g.अतः f = h \circ g।Isliye f = h \circ g.(h \circ g)(x) = h(1 - x + |x|) = |1 - x + |x|| = f(x)
- Sum of continuous functions is continuous.संतत फलनों का योग संतत है।Continuous functions ka sum continuous hai.h \text{ continuous (Ex. 7)},\ g = \text{polynomial} + |x| \text{ continuous}
✔ f, a composite of two continuous functions, is continuous.f, दो संतत फलनों का संयुक्त होने से, संतत है।f, do continuous functions ka composite hone se, continuous hai.
(Exercise 5.1, Q5) Is f(x) = \begin{cases} x, & x \le 1 \\ 5, & x > 1 \end{cases} continuous at x = 0? At x = 1? At x = 2?(प्रश्नावली 5.1, प्र. 5) क्या f(x) = \begin{cases} x, & x \le 1 \\ 5, & x > 1 \end{cases}, x = 0 पर संतत है? x = 1 पर? x = 2 पर?(Exercise 5.1, Q5) Kya f(x) = \begin{cases} x, & x \le 1 \\ 5, & x > 1 \end{cases}, x = 0 par continuous hai? x = 1 par? x = 2 par?
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f is continuous at x = 0 and x = 2; not continuous at x = 1 (LHL = 1, RHL = 5).f, x = 0 और x = 2 पर संतत है; x = 1 पर संतत नहीं (LHL = 1, RHL = 5)।f, x = 0 aur x = 2 par continuous hai; x = 1 par continuous nahi (LHL = 1, RHL = 5).
(Exercise 5.1, Q8) Find all points of discontinuity of f(x) = \begin{cases} \dfrac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}(प्रश्नावली 5.1, प्र. 8) f(x) = \begin{cases} \dfrac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} के सभी असांतत्य के बिंदु ज्ञात कीजिए।(Exercise 5.1, Q8) f(x) = \begin{cases} \dfrac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} ke saare points of discontinuity nikaaliye.
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Discontinuous at x = 0 (LHL = -1, RHL = 1).x = 0 पर असंतत (LHL = -1, RHL = 1)।x = 0 par discontinuous (LHL = -1, RHL = 1).
(Exercise 5.1, Q17) Find the relationship between a and b so that f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases} is continuous at x = 3.(प्रश्नावली 5.1, प्र. 17) a और b में संबंध ज्ञात कीजिए ताकि f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases}, x = 3 पर संतत हो।(Exercise 5.1, Q17) a aur b mein relation nikaaliye taaki f(x) = \begin{cases} ax + 1, & x \le 3 \\ bx + 3, & x > 3 \end{cases}, x = 3 par continuous ho.
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3a + 1 = 3b + 3, i.e., a = b + \dfrac{2}{3}.3a + 1 = 3b + 3, अर्थात् a = b + \dfrac{2}{3}।3a + 1 = 3b + 3, yaani a = b + \dfrac{2}{3}.
(Exercise 5.1, Q18) For what value of λ is f(x) = \begin{cases} \lambda(x^2 - 2x), & x \le 0 \\ 4x + 1, & x > 0 \end{cases} continuous at x = 0? What about x = 1?(प्रश्नावली 5.1, प्र. 18) λ के किस मान के लिए f(x) = \begin{cases} \lambda(x^2 - 2x), & x \le 0 \\ 4x + 1, & x > 0 \end{cases}, x = 0 पर संतत है? x = 1 पर क्या?(Exercise 5.1, Q18) λ ki kis value ke liye f(x) = \begin{cases} \lambda(x^2 - 2x), & x \le 0 \\ 4x + 1, & x > 0 \end{cases}, x = 0 par continuous hai? x = 1 par?
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For no value of λ is f continuous at x = 0 (LHL = 0, RHL = 1), but f is continuous at x = 1 for any value of λ.λ के किसी भी मान के लिए f, x = 0 पर संतत नहीं (LHL = 0, RHL = 1), पर x = 1 पर λ के हर मान के लिए संतत है।λ ki kisi bhi value ke liye f, x = 0 par continuous nahi (LHL = 0, RHL = 1), par x = 1 par λ ki har value ke liye continuous hai.
(Exercise 5.1, Q26) Find k so that f(x) = \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & x \ne \dfrac{\pi}{2} \\ 3, & x = \dfrac{\pi}{2} \end{cases} is continuous at x = \dfrac{\pi}{2}.(प्रश्नावली 5.1, प्र. 26) k ज्ञात कीजिए ताकि f(x) = \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & x \ne \dfrac{\pi}{2} \\ 3, & x = \dfrac{\pi}{2} \end{cases}, x = \dfrac{\pi}{2} पर संतत हो।(Exercise 5.1, Q26) k nikaaliye taaki f(x) = \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & x \ne \dfrac{\pi}{2} \\ 3, & x = \dfrac{\pi}{2} \end{cases}, x = \dfrac{\pi}{2} par continuous ho.
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k = 6 (the limit is \dfrac{k}{2}, and \dfrac{k}{2} = 3).k = 6 (सीमा \dfrac{k}{2} है, और \dfrac{k}{2} = 3)।k = 6 (limit \dfrac{k}{2} hai, aur \dfrac{k}{2} = 3).
(Exercise 5.1, Q27) Find k so that f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases} is continuous at x = 2.(प्रश्नावली 5.1, प्र. 27) k ज्ञात कीजिए ताकि f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases}, x = 2 पर संतत हो।(Exercise 5.1, Q27) k nikaaliye taaki f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases}, x = 2 par continuous ho.
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k = \dfrac{3}{4}k = \dfrac{3}{4}k = \dfrac{3}{4}
(Exercise 5.1, Q28) Find k so that f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases} is continuous at x = \pi.(प्रश्नावली 5.1, प्र. 28) k ज्ञात कीजिए ताकि f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}, x = \pi पर संतत हो।(Exercise 5.1, Q28) k nikaaliye taaki f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}, x = \pi par continuous ho.
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k = -\dfrac{2}{\pi}k = -\dfrac{2}{\pi}k = -\dfrac{2}{\pi}
(Exercise 5.1, Q30) Find a and b such that f(x) = \begin{cases} 5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10 \end{cases} is a continuous function.(प्रश्नावली 5.1, प्र. 30) a और b ज्ञात कीजिए ताकि f(x) = \begin{cases} 5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10 \end{cases} एक संतत फलन हो।(Exercise 5.1, Q30) a aur b nikaaliye taaki f(x) = \begin{cases} 5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10 \end{cases} ek continuous function ho.
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a = 2,\ b = 1 (from 2a + b = 5 and 10a + b = 21).a = 2,\ b = 1 (2a + b = 5 और 10a + b = 21 से)।a = 2,\ b = 1 (2a + b = 5 aur 10a + b = 21 se).
(Exercise 5.1, Q34) Find all the points of discontinuity of f(x) = |x| - |x + 1|.(प्रश्नावली 5.1, प्र. 34) f(x) = |x| - |x + 1| के सभी असांतत्य के बिंदु ज्ञात कीजिए।(Exercise 5.1, Q34) f(x) = |x| - |x + 1| ke saare points of discontinuity nikaaliye.
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There is no point of discontinuity.कोई असांतत्य का बिंदु नहीं है।Koi point of discontinuity nahi hai.
5.3DifferentiabilityअवकलनीयताDifferentiability
Secant → tangent: the derivative is a limit of slopesछेदक रेखा → स्पर्श रेखा: अवकलज ढालों की सीमा हैSecant → tangent: derivative slopes ki limit haiNCERT Section 5.3 · 10 tappable partsOpenAlgebra of derivatives: (1) (u \pm v)' = u' \pm v' (2) (uv)' = u'v + uv' (Leibnitz or product rule) (3) \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}, wherever v \ne 0 (quotient rule). Table 5.3: \dfrac{d}{dx}x^n = nx^{n-1}, \dfrac{d}{dx}\sin x = \cos x, \dfrac{d}{dx}\cos x = -\sin x, \dfrac{d}{dx}\tan x = \sec^2 x.अवकलजों का बीजगणित: (1) (u \pm v)' = u' \pm v' (2) (uv)' = u'v + uv' (लाइबनिट्ज़ या गुणनफल नियम) (3) \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}, जहाँ v \ne 0 (भागफल नियम)। सारणी 5.3: \dfrac{d}{dx}x^n = nx^{n-1}, \dfrac{d}{dx}\sin x = \cos x, \dfrac{d}{dx}\cos x = -\sin x, \dfrac{d}{dx}\tan x = \sec^2 x।Derivatives ka algebra: (1) (u \pm v)' = u' \pm v' (2) (uv)' = u'v + uv' (Leibnitz ya product rule) (3) \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}, jahan v \ne 0 (quotient rule). Table 5.3: \dfrac{d}{dx}x^n = nx^{n-1}, \dfrac{d}{dx}\sin x = \cos x, \dfrac{d}{dx}\cos x = -\sin x, \dfrac{d}{dx}\tan x = \sec^2 x.Sum, product and quotient rules, and the basic derivatives table.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Given: f is differentiable at c.दिया है: f, c पर अवकलनीय है।Diya hai: f, c par differentiable hai.\lim_{x \to c} \dfrac{f(x) - f(c)}{x - c} = f'(c)
- Multiply and divide by x - c.x - c से गुणा और भाग।x - c se multiply aur divide.f(x) - f(c) = \dfrac{f(x) - f(c)}{x - c} \cdot (x - c),\quad x \ne c
- Limit of a product = product of limits.गुणनफल की सीमा = सीमाओं का गुणनफल।Product ki limit = limits ka product.\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \dfrac{f(x) - f(c)}{x - c} \cdot \lim_{x \to c} (x - c) = f'(c) \cdot 0 = 0
✔ \lim_{x \to c} f(x) = f(c), so f is continuous at c. Corollary 1: every differentiable function is continuous.\lim_{x \to c} f(x) = f(c), अतः f, c पर संतत है। उपप्रमेय 1: प्रत्येक अवकलनीय फलन संतत होता है।\lim_{x \to c} f(x) = f(c), isliye f, c par continuous hai. Corollary 1: har differentiable function continuous hota hai.
The converse is NOT true. f(x) = |x| is continuous (Example 7), but at 0: left hand limit \lim_{h \to 0^-} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{-h}{h} = -1 and right hand limit \lim_{h \to 0^+} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{h}{h} = 1. They are not equal, so f'(0) does not exist: |x| is not differentiable at 0, hence not a differentiable function. On the graph you see a sharp corner at 0.इसका विलोम सत्य नहीं है। f(x) = |x| संतत है (उदाहरण 7), पर 0 पर: बाएँ पक्ष की सीमा \lim_{h \to 0^-} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{-h}{h} = -1 और दाएँ पक्ष की सीमा \lim_{h \to 0^+} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{h}{h} = 1। ये बराबर नहीं, अतः f'(0) विद्यमान नहीं: |x|, 0 पर अवकलनीय नहीं है, इसलिए अवकलनीय फलन नहीं है। ग्राफ़ पर 0 पर नुकीला कोना दिखता है।Iska converse sach nahi hai. f(x) = |x| continuous hai (Example 7), par 0 par: left hand limit \lim_{h \to 0^-} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{-h}{h} = -1 aur right hand limit \lim_{h \to 0^+} \dfrac{f(0 + h) - f(0)}{h} = \dfrac{h}{h} = 1. Ye barabar nahi, isliye f'(0) exist nahi karta: |x|, 0 par differentiable nahi hai, isliye differentiable function nahi hai. Graph par 0 par nukila kona dikhta hai.|x| is continuous at 0 but has a corner there (slopes −1 and 1), so it is not differentiable at 0.
f(x) = |x|: continuous at 0 but not differentiable theref(x) = |x|: 0 पर संतत पर अवकलनीय नहींf(x) = |x|: 0 par continuous par differentiable nahiNCERT Section 5.3 · 6 tappable partsOpen5.3.1Derivatives of composite functionsसंयुक्त फलनों के अवकलजComposite functions ke derivatives
Take f(x) = (2x + 1)^3. Expanding: \dfrac{d}{dx}(8x^3 + 12x^2 + 6x + 1) = 24x^2 + 24x + 6 = 6(2x + 1)^2. Now see it as a box inside a box: inner t = g(x) = 2x + 1, outer h(t) = t^3. Then \dfrac{dh}{dt} \cdot \dfrac{dt}{dx} = 3t^2 \cdot 2 = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2 — the same answer, without expanding. This saves huge work for, say, (2x + 1)^{100}.f(x) = (2x + 1)^3 लीजिए। प्रसार करके: \dfrac{d}{dx}(8x^3 + 12x^2 + 6x + 1) = 24x^2 + 24x + 6 = 6(2x + 1)^2। अब इसे डिब्बे के अंदर डिब्बे की तरह देखिए: भीतर t = g(x) = 2x + 1, बाहर h(t) = t^3। तब \dfrac{dh}{dt} \cdot \dfrac{dt}{dx} = 3t^2 \cdot 2 = 6(2x + 1)^2 — वही उत्तर, बिना प्रसार के। (2x + 1)^{100} जैसे फलन के लिए यह बहुत मेहनत बचाता है।f(x) = (2x + 1)^3 lijiye. Expand karke: \dfrac{d}{dx}(8x^3 + 12x^2 + 6x + 1) = 24x^2 + 24x + 6 = 6(2x + 1)^2. Ab ise dabbe ke andar dabba ki tarah dekhiye: andar t = g(x) = 2x + 1, bahar h(t) = t^3. Tab \dfrac{dh}{dt} \cdot \dfrac{dt}{dx} = 3t^2 \cdot 2 = 6(2x + 1)^2 — wahi answer, bina expand kiye. (2x + 1)^{100} jaise function ke liye ye bahut mehnat bachata hai.Outer derivative × inner derivative gives the same answer as expanding.
Chain rule: a box inside a boxश्रृंखला नियम: डिब्बे के अंदर डिब्बाChain rule: dabbe ke andar dabbaNCERT Theorem 4 · 7 tappable partsOpenSolution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Inner box x^2, outer box sin.भीतरी डिब्बा x^2, बाहरी डिब्बा sin।Andar ka dabba x^2, bahar ka dabba sin.t = u(x) = x^2,\ v(t) = \sin t,\ f = v \circ u
- Both derivatives exist.दोनों अवकलज विद्यमान हैं।Dono derivatives exist karte hain.\dfrac{dv}{dt} = \cos t,\quad \dfrac{dt}{dx} = 2x
- Chain rule.श्रृंखला नियम।Chain rule.\dfrac{df}{dx} = \cos t \cdot 2x
✔ \dfrac{df}{dx} = 2x \cos x^2 (final answer written in terms of x).\dfrac{df}{dx} = 2x \cos x^2 (अंतिम उत्तर x के पदों में)।\dfrac{df}{dx} = 2x \cos x^2 (final answer x mein likhiye).
(Exercise 5.2, Q1) Differentiate \sin(x^2 + 5) w.r.t. x.(प्रश्नावली 5.2, प्र. 1) \sin(x^2 + 5) का x के सापेक्ष अवकलन कीजिए।(Exercise 5.2, Q1) \sin(x^2 + 5) ko x ke respect mein differentiate kijiye.
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2x \cos(x^2 + 5)2x \cos(x^2 + 5)2x \cos(x^2 + 5)
(Exercise 5.2, Q2) Differentiate \cos(\sin x) w.r.t. x.(प्रश्नावली 5.2, प्र. 2) \cos(\sin x) का x के सापेक्ष अवकलन कीजिए।(Exercise 5.2, Q2) \cos(\sin x) ko x ke respect mein differentiate kijiye.
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-\cos x \sin(\sin x)-\cos x \sin(\sin x)-\cos x \sin(\sin x)
(Exercise 5.2, Q3) Differentiate \sin(ax + b) w.r.t. x.(प्रश्नावली 5.2, प्र. 3) \sin(ax + b) का x के सापेक्ष अवकलन कीजिए।(Exercise 5.2, Q3) \sin(ax + b) ko x ke respect mein differentiate kijiye.
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a \cos(ax + b)a \cos(ax + b)a \cos(ax + b)
(Exercise 5.2, Q9) Prove that f(x) = |x - 1|, x \in \mathbf{R}, is not differentiable at x = 1.(प्रश्नावली 5.2, प्र. 9) सिद्ध कीजिए कि f(x) = |x - 1|, x \in \mathbf{R}, x = 1 पर अवकलनीय नहीं है।(Exercise 5.2, Q9) Prove kijiye ki f(x) = |x - 1|, x \in \mathbf{R}, x = 1 par differentiable nahi hai.
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At x = 1: left hand derivative = -1, right hand derivative = 1; they are unequal, so f is not differentiable at x = 1.x = 1 पर बायाँ अवकलज = -1, दायाँ अवकलज = 1; ये बराबर नहीं, अतः f, x = 1 पर अवकलनीय नहीं।x = 1 par left derivative = -1, right derivative = 1; ye barabar nahi, isliye f, x = 1 par differentiable nahi.
5.3.2Derivatives of implicit functionsअस्पष्ट फलनों के अवकलजImplicit functions ke derivatives
If a relation can easily be solved for y as y = f(x) (like x - y - \pi = 0 \Rightarrow y = x - \pi), y is an explicit function of x. If not (like x + \sin xy - y = 0), y still depends on x, and the relation gives y implicitly. To differentiate: differentiate both sides w.r.t. x, treat y as a function of x (so \dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx} by the chain rule), then solve for \dfrac{dy}{dx}.यदि किसी संबंध को आसानी से y = f(x) के रूप में हल किया जा सके (जैसे x - y - \pi = 0 \Rightarrow y = x - \pi), तो y, x का स्पष्ट फलन है। यदि नहीं (जैसे x + \sin xy - y = 0), तब भी y, x पर निर्भर है और संबंध y को अस्पष्ट रूप से देता है। अवकलन के लिए: दोनों पक्षों का x के सापेक्ष अवकलन कीजिए, y को x का फलन मानिए (अतः श्रृंखला नियम से \dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}), फिर \dfrac{dy}{dx} के लिए हल कीजिए।Agar kisi relation ko aasaani se y = f(x) ki tarah solve kar sakein (jaise x - y - \pi = 0 \Rightarrow y = x - \pi), to y, x ka explicit function hai. Agar nahi (jaise x + \sin xy - y = 0), tab bhi y, x par depend karta hai aur relation y ko implicitly deta hai. Differentiate karne ke liye: dono sides ko x ke respect mein differentiate kijiye, y ko x ka function maaniye (to chain rule se \dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}), phir \dfrac{dy}{dx} ke liye solve kijiye.Implicit: differentiate both sides, multiply y-terms' derivatives by dy/dx, solve for dy/dx.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Way 1: solve for y first.तरीका 1: पहले y निकालिए।Tareeka 1: pehle y nikaaliye.y = x - \pi \Rightarrow \dfrac{dy}{dx} = 1
- Way 2: differentiate directly; π is a constant, so its derivative is 0.तरीका 2: सीधे अवकलन; π अचर है, इसलिए उसका अवकलज 0।Tareeka 2: seedha differentiate; π constant hai, isliye uska derivative 0.\dfrac{d}{dx}(x - y) = \dfrac{d\pi}{dx} \Rightarrow \dfrac{d}{dx}(x) - \dfrac{d}{dx}(y) = 0
✔ \dfrac{dy}{dx} = \dfrac{dx}{dx} = 1\dfrac{dy}{dx} = \dfrac{dx}{dx} = 1\dfrac{dy}{dx} = \dfrac{dx}{dx} = 1
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Differentiate both sides w.r.t. x.दोनों पक्षों का x के सापेक्ष अवकलन।Dono sides ko x ke respect mein differentiate kijiye.\dfrac{dy}{dx} + \dfrac{d}{dx}(\sin y) = \dfrac{d}{dx}(\cos x)
- Chain rule: \dfrac{d}{dx}\sin y = \cos y \dfrac{dy}{dx}.श्रृंखला नियम: \dfrac{d}{dx}\sin y = \cos y \dfrac{dy}{dx}।Chain rule: \dfrac{d}{dx}\sin y = \cos y \dfrac{dy}{dx}.\dfrac{dy}{dx} + \cos y \cdot \dfrac{dy}{dx} = -\sin x
- Take \dfrac{dy}{dx} common.\dfrac{dy}{dx} उभयनिष्ठ लीजिए।\dfrac{dy}{dx} common lijiye.(1 + \cos y)\dfrac{dy}{dx} = -\sin x
✔ \dfrac{dy}{dx} = -\dfrac{\sin x}{1 + \cos y}, where y \ne (2n + 1)\pi\dfrac{dy}{dx} = -\dfrac{\sin x}{1 + \cos y}, जहाँ y \ne (2n + 1)\pi\dfrac{dy}{dx} = -\dfrac{\sin x}{1 + \cos y}, jahan y \ne (2n + 1)\pi
(Exercise 5.3, Q1) Find \dfrac{dy}{dx}: 2x + 3y = \sin x(प्रश्नावली 5.3, प्र. 1) \dfrac{dy}{dx} ज्ञात कीजिए: 2x + 3y = \sin x(Exercise 5.3, Q1) \dfrac{dy}{dx} nikaaliye: 2x + 3y = \sin x
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\dfrac{\cos x - 2}{3}\dfrac{\cos x - 2}{3}\dfrac{\cos x - 2}{3}
(Exercise 5.3, Q2) Find \dfrac{dy}{dx}: 2x + 3y = \sin y(प्रश्नावली 5.3, प्र. 2) \dfrac{dy}{dx} ज्ञात कीजिए: 2x + 3y = \sin y(Exercise 5.3, Q2) \dfrac{dy}{dx} nikaaliye: 2x + 3y = \sin y
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\dfrac{2}{\cos y - 3}\dfrac{2}{\cos y - 3}\dfrac{2}{\cos y - 3}
(Exercise 5.3, Q5) Find \dfrac{dy}{dx}: x^2 + xy + y^2 = 100(प्रश्नावली 5.3, प्र. 5) \dfrac{dy}{dx} ज्ञात कीजिए: x^2 + xy + y^2 = 100(Exercise 5.3, Q5) \dfrac{dy}{dx} nikaaliye: x^2 + xy + y^2 = 100
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-\dfrac{2x + y}{x + 2y}-\dfrac{2x + y}{x + 2y}-\dfrac{2x + y}{x + 2y}
(Exercise 5.3, Q8) Find \dfrac{dy}{dx}: \sin^2 x + \cos^2 y = 1(प्रश्नावली 5.3, प्र. 8) \dfrac{dy}{dx} ज्ञात कीजिए: \sin^2 x + \cos^2 y = 1(Exercise 5.3, Q8) \dfrac{dy}{dx} nikaaliye: \sin^2 x + \cos^2 y = 1
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\dfrac{\sin 2x}{\sin 2y}\dfrac{\sin 2x}{\sin 2y}\dfrac{\sin 2x}{\sin 2y}
5.3.3Derivatives of inverse trigonometric functionsप्रतिलोम त्रिकोणमितीय फलनों के अवकलजInverse trigonometric functions ke derivatives
Inverse trigonometric functions are continuous functions (we do not prove this). We find their derivatives with the chain rule: write y = \sin^{-1} x as x = \sin y and differentiate implicitly.प्रतिलोम त्रिकोणमितीय फलन संतत फलन हैं (इसे हम सिद्ध नहीं करेंगे)। इनके अवकलज श्रृंखला नियम से निकालते हैं: y = \sin^{-1} x को x = \sin y लिखकर अस्पष्ट अवकलन कीजिए।Inverse trigonometric functions continuous functions hain (iska proof nahi karenge). Inke derivatives chain rule se nikaalte hain: y = \sin^{-1} x ko x = \sin y likhkar implicit differentiation kijiye.Write y = sin⁻¹x as x = sin y and differentiate.
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Rewrite.पुनः लिखिए।Dobara likhiye.y = \sin^{-1} x \Rightarrow x = \sin y
- Differentiate both sides w.r.t. x; defined only for \cos y \ne 0, i.e., x \ne -1, 1.दोनों पक्षों का x के सापेक्ष अवकलन; केवल \cos y \ne 0, अर्थात् x \ne -1, 1 के लिए परिभाषित।Dono sides ko x ke respect mein differentiate kijiye; sirf \cos y \ne 0, yaani x \ne -1, 1 ke liye defined.1 = \cos y \dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{\cos y}
- Since \sin(\sin^{-1} x) = x for x \in (-1, 1).क्योंकि x \in (-1, 1) के लिए \sin(\sin^{-1} x) = x।Kyunki x \in (-1, 1) ke liye \sin(\sin^{-1} x) = x.\cos^2 y = 1 - \sin^2 y = 1 - x^2
- Choose the positive square root (principal value branch).धनात्मक वर्गमूल लीजिए (मुख्य मान शाखा)।Positive square root lijiye (principal value branch).y \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \Rightarrow \cos y > 0 \Rightarrow \cos y = \sqrt{1 - x^2}
✔ \dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}, for x \in (-1, 1).\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}, x \in (-1, 1) के लिए।\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}, x \in (-1, 1) ke liye.
(Exercise 5.3, Q10) Find \dfrac{dy}{dx}: y = \tan^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right),\ -\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}(प्रश्नावली 5.3, प्र. 10) \dfrac{dy}{dx} ज्ञात कीजिए: y = \tan^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right),\ -\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}(Exercise 5.3, Q10) \dfrac{dy}{dx} nikaaliye: y = \tan^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right),\ -\dfrac{1}{\sqrt{3}} < x < \dfrac{1}{\sqrt{3}}
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\dfrac{3}{1 + x^2}\dfrac{3}{1 + x^2}\dfrac{3}{1 + x^2}
(Exercise 5.3, Q13) Find \dfrac{dy}{dx}: y = \cos^{-1}\left(\dfrac{2x}{1 + x^2}\right),\ -1 < x < 1(प्रश्नावली 5.3, प्र. 13) \dfrac{dy}{dx} ज्ञात कीजिए: y = \cos^{-1}\left(\dfrac{2x}{1 + x^2}\right),\ -1 < x < 1(Exercise 5.3, Q13) \dfrac{dy}{dx} nikaaliye: y = \cos^{-1}\left(\dfrac{2x}{1 + x^2}\right),\ -1 < x < 1
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\dfrac{-2}{1 + x^2}\dfrac{-2}{1 + x^2}\dfrac{-2}{1 + x^2}
5.4Exponential and Logarithmic Functionsचरघातांकी तथा लघुगणकीय फलनExponential aur Logarithmic Functions
Growth. The graphs of x, x^2, x^3, x^4 (Fig 5.9) get steeper as the power grows: from x = 1 to x = 2, x^{10} goes from 1 to 2^{10} while x^{15} goes from 1 to 2^{15}. But y = 10^x grows faster than any x^n: at x = 10^3, x^{100} = 10^{300} while 10^x = 10^{1000}.वृद्धि। x, x^2, x^3, x^4 के ग्राफ़ (आकृति 5.9) घात बढ़ने पर अधिक खड़े होते जाते हैं: x = 1 से x = 2 तक x^{10}, 1 से 2^{10} तक जाता है जबकि x^{15}, 1 से 2^{15} तक। पर y = 10^x किसी भी x^n से तेज़ बढ़ता है: x = 10^3 पर x^{100} = 10^{300} जबकि 10^x = 10^{1000}।Growth. x, x^2, x^3, x^4 ke graph (Fig 5.9) power badhne par zyada khade hote jaate hain: x = 1 se x = 2 tak x^{10}, 1 se 2^{10} tak jaata hai jabki x^{15}, 1 se 2^{15} tak. Par y = 10^x kisi bhi x^n se tez badhta hai: x = 10^3 par x^{100} = 10^{300} jabki 10^x = 10^{1000}.10^x beats every power of x for large x.
y = eˣ and y = log x: mirror images in y = xy = eˣ और y = log x: y = x में दर्पण प्रतिबिंबy = eˣ aur y = log x: y = x mein mirror imageNCERT Fig 5.11 · 8 tappable partsOpenTwo properties. (1) Change of base: let \log_a p = \alpha, \log_b p = \beta, \log_b a = \gamma, i.e., a^\alpha = p, b^\beta = p, b^\gamma = a. Then (b^\gamma)^\alpha = b^{\gamma\alpha} = p = b^\beta, so \beta = \alpha\gamma, giving \log_a p = \dfrac{\log_b p}{\log_b a}. (2) Product: if \log_b pq = \alpha, \log_b p = \beta, \log_b q = \gamma, then b^\alpha = pq = b^\beta b^\gamma = b^{\beta + \gamma}, so \log_b pq = \log_b p + \log_b q. Consequences: \log_b p^n = n \log_b p (true for any real n) and \log_b \dfrac{x}{y} = \log_b x - \log_b y.दो गुणधर्म। (1) आधार परिवर्तन: मान लीजिए \log_a p = \alpha, \log_b p = \beta, \log_b a = \gamma, अर्थात् a^\alpha = p, b^\beta = p, b^\gamma = a। तब (b^\gamma)^\alpha = b^{\gamma\alpha} = p = b^\beta, अतः \beta = \alpha\gamma, जिससे \log_a p = \dfrac{\log_b p}{\log_b a}। (2) गुणनफल: यदि \log_b pq = \alpha, \log_b p = \beta, \log_b q = \gamma, तो b^\alpha = pq = b^\beta b^\gamma = b^{\beta + \gamma}, अतः \log_b pq = \log_b p + \log_b q। परिणाम: \log_b p^n = n \log_b p (किसी भी वास्तविक n के लिए) और \log_b \dfrac{x}{y} = \log_b x - \log_b y।Do properties. (1) Change of base: maan lijiye \log_a p = \alpha, \log_b p = \beta, \log_b a = \gamma, yaani a^\alpha = p, b^\beta = p, b^\gamma = a. Tab (b^\gamma)^\alpha = b^{\gamma\alpha} = p = b^\beta, to \beta = \alpha\gamma, jisse \log_a p = \dfrac{\log_b p}{\log_b a}. (2) Product: agar \log_b pq = \alpha, \log_b p = \beta, \log_b q = \gamma, to b^\alpha = pq = b^\beta b^\gamma = b^{\beta + \gamma}, isliye \log_b pq = \log_b p + \log_b q. Nateeje: \log_b p^n = n \log_b p (kisi bhi real n ke liye) aur \log_b \dfrac{x}{y} = \log_b x - \log_b y.Change of base, log of a product = sum of logs, log of a power = power × log.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- For x \le 0, \log x is not defined, so the equation is not true there.x \le 0 के लिए \log x परिभाषित नहीं, इसलिए वहाँ समीकरण सत्य नहीं।x \le 0 ke liye \log x defined nahi, isliye wahan equation sach nahi.\text{Domain of } \log = \mathbf{R}^+
- For x > 0, take log of both sides (\log e = 1).x > 0 के लिए दोनों पक्षों का लघुगणक लीजिए (\log e = 1)।x > 0 ke liye dono sides ka log lijiye (\log e = 1).y = e^{\log x} > 0 \Rightarrow \log y = \log(e^{\log x}) = \log x \cdot \log e = \log x
- log is one-one (ever increasing).log एकैकी है (सदैव वर्धमान)।log one-one hai (hamesha increasing).\log y = \log x \Rightarrow y = x
✔ No. x = e^{\log x} is true only for positive values of x.नहीं। x = e^{\log x} केवल x के धनात्मक मानों के लिए सत्य है।Nahi. x = e^{\log x} sirf x ki positive values ke liye sach hai.
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Chain rule: outer e^t, inner -x.श्रृंखला नियम: बाहर e^t, भीतर -x।Chain rule: bahar e^t, andar -x.\text{(i)}\ \dfrac{d}{dx}e^{-x} = e^{-x} \cdot \dfrac{d}{dx}(-x) = -e^{-x}
- Outer sin, inner log x.बाहर sin, भीतर log x।Bahar sin, andar log x.\text{(ii)}\ \dfrac{d}{dx}\sin(\log x) = \cos(\log x) \cdot \dfrac{1}{x} = \dfrac{\cos(\log x)}{x}
- Outer \cos^{-1}, inner e^x.बाहर \cos^{-1}, भीतर e^x।Bahar \cos^{-1}, andar e^x.\text{(iii)}\ \dfrac{d}{dx}\cos^{-1}(e^x) = \dfrac{-1}{\sqrt{1 - (e^x)^2}} \cdot e^x = \dfrac{-e^x}{\sqrt{1 - e^{2x}}}
- Outer e^t, inner cos x.बाहर e^t, भीतर cos x।Bahar e^t, andar cos x.\text{(iv)}\ \dfrac{d}{dx}e^{\cos x} = e^{\cos x} \cdot (-\sin x)
✔ (i) -e^{-x} (ii) \dfrac{\cos(\log x)}{x} (iii) \dfrac{-e^x}{\sqrt{1 - e^{2x}}} (iv) -(\sin x)\, e^{\cos x}(i) -e^{-x} (ii) \dfrac{\cos(\log x)}{x} (iii) \dfrac{-e^x}{\sqrt{1 - e^{2x}}} (iv) -(\sin x)\, e^{\cos x}(i) -e^{-x} (ii) \dfrac{\cos(\log x)}{x} (iii) \dfrac{-e^x}{\sqrt{1 - e^{2x}}} (iv) -(\sin x)\, e^{\cos x}
(Exercise 5.4, Q1) Differentiate \dfrac{e^x}{\sin x} w.r.t. x.(प्रश्नावली 5.4, प्र. 1) \dfrac{e^x}{\sin x} का x के सापेक्ष अवकलन कीजिए।(Exercise 5.4, Q1) \dfrac{e^x}{\sin x} ko x ke respect mein differentiate kijiye.
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\dfrac{e^x(\sin x - \cos x)}{\sin^2 x},\ x \ne n\pi,\ n \in \mathbf{Z}\dfrac{e^x(\sin x - \cos x)}{\sin^2 x},\ x \ne n\pi,\ n \in \mathbf{Z}\dfrac{e^x(\sin x - \cos x)}{\sin^2 x},\ x \ne n\pi,\ n \in \mathbf{Z}
(Exercise 5.4, Q2) Differentiate e^{\sin^{-1} x} w.r.t. x.(प्रश्नावली 5.4, प्र. 2) e^{\sin^{-1} x} का x के सापेक्ष अवकलन कीजिए।(Exercise 5.4, Q2) e^{\sin^{-1} x} ko x ke respect mein differentiate kijiye.
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\dfrac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}},\ x \in (-1, 1)\dfrac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}},\ x \in (-1, 1)\dfrac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}},\ x \in (-1, 1)
(Exercise 5.4, Q3) Differentiate e^{x^3} w.r.t. x.(प्रश्नावली 5.4, प्र. 3) e^{x^3} का x के सापेक्ष अवकलन कीजिए।(Exercise 5.4, Q3) e^{x^3} ko x ke respect mein differentiate kijiye.
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3x^2 e^{x^3}3x^2 e^{x^3}3x^2 e^{x^3}
(Exercise 5.4, Q5) Differentiate \log(\cos e^x) w.r.t. x.(प्रश्नावली 5.4, प्र. 5) \log(\cos e^x) का x के सापेक्ष अवकलन कीजिए।(Exercise 5.4, Q5) \log(\cos e^x) ko x ke respect mein differentiate kijiye.
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-e^x \tan e^x,\ e^x \ne (2n + 1)\dfrac{\pi}{2},\ n \in \mathbf{N}-e^x \tan e^x,\ e^x \ne (2n + 1)\dfrac{\pi}{2},\ n \in \mathbf{N}-e^x \tan e^x,\ e^x \ne (2n + 1)\dfrac{\pi}{2},\ n \in \mathbf{N}
(Exercise 5.4, Q8) Differentiate \log(\log x),\ x > 1 w.r.t. x.(प्रश्नावली 5.4, प्र. 8) \log(\log x),\ x > 1 का x के सापेक्ष अवकलन कीजिए।(Exercise 5.4, Q8) \log(\log x),\ x > 1 ko x ke respect mein differentiate kijiye.
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\dfrac{1}{x \log x},\ x > 1\dfrac{1}{x \log x},\ x > 1\dfrac{1}{x \log x},\ x > 1
5.5Logarithmic Differentiationलघुगणकीय अवकलनLogarithmic Differentiation
For y = f(x) = [u(x)]^{v(x)}, taking logarithm (to base e): \log y = v(x) \log[u(x)]. By the chain rule, \dfrac{1}{y} \cdot \dfrac{dy}{dx} = v(x) \cdot \dfrac{1}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)], so \frac{dy}{dx} = y\left[\frac{v(x)}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)]\right] The main point: f(x) and u(x) must always be positive, otherwise their logarithms are not defined. This process is called logarithmic differentiation.y = f(x) = [u(x)]^{v(x)} के लिए लघुगणक (आधार e) लेने पर: \log y = v(x) \log[u(x)]। श्रृंखला नियम से \dfrac{1}{y} \cdot \dfrac{dy}{dx} = v(x) \cdot \dfrac{1}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)], अतः \frac{dy}{dx} = y\left[\frac{v(x)}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)]\right] मुख्य बात: f(x) और u(x) हमेशा धनात्मक होने चाहिए, नहीं तो उनके लघुगणक परिभाषित नहीं। इस प्रक्रिया को लघुगणकीय अवकलन कहते हैं।y = f(x) = [u(x)]^{v(x)} ke liye log (base e) lene par: \log y = v(x) \log[u(x)]. Chain rule se \dfrac{1}{y} \cdot \dfrac{dy}{dx} = v(x) \cdot \dfrac{1}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)], isliye \frac{dy}{dx} = y\left[\frac{v(x)}{u(x)} \cdot u'(x) + v'(x) \cdot \log[u(x)]\right] Main baat: f(x) aur u(x) hamesha positive hone chahiye, warna unke log defined nahi. Is process ko logarithmic differentiation kehte hain.For y = u^v: log y = v log u, differentiate, multiply by y. u and f must be positive.
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Name the function y.फलन को y कहिए।Function ko y kahiye.y = \sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}
- Take log: root → \frac{1}{2}, product → sum, quotient → difference.लघुगणक लीजिए: वर्गमूल → \frac{1}{2}, गुणनफल → योग, भागफल → अंतर।Log lijiye: root → \frac{1}{2}, product → sum, quotient → difference.\log y = \dfrac{1}{2}\left[\log(x - 3) + \log(x^2 + 4) - \log(3x^2 + 4x + 5)\right]
- Differentiate both sides w.r.t. x.दोनों पक्षों का x के सापेक्ष अवकलन।Dono sides ko x ke respect mein differentiate kijiye.\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{2}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]
- Multiply by y.y से गुणा कीजिए।y se multiply kijiye.\dfrac{dy}{dx} = \dfrac{y}{2}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]
✔ \dfrac{dy}{dx} = \dfrac{1}{2}\sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]\dfrac{dy}{dx} = \dfrac{1}{2}\sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]\dfrac{dy}{dx} = \dfrac{1}{2}\sqrt{\dfrac{(x - 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[\dfrac{1}{x - 3} + \dfrac{2x}{x^2 + 4} - \dfrac{6x + 4}{3x^2 + 4x + 5}\right]
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Take log; \log a is just a constant.लघुगणक लीजिए; \log a केवल एक अचर है।Log lijiye; \log a sirf ek constant hai.y = a^x \Rightarrow \log y = x \log a
- Differentiate both sides.दोनों पक्षों का अवकलन।Dono sides differentiate kijiye.\dfrac{1}{y}\dfrac{dy}{dx} = \log a \Rightarrow \dfrac{dy}{dx} = y \log a
- Write a^x = e^{x \log a} and use the chain rule.a^x = e^{x \log a} लिखकर श्रृंखला नियम।a^x = e^{x \log a} likhkar chain rule.\text{Alternatively: } \dfrac{d}{dx}(a^x) = \dfrac{d}{dx}(e^{x \log a}) = e^{x \log a} \cdot \log a
✔ \dfrac{d}{dx}(a^x) = a^x \log a\dfrac{d}{dx}(a^x) = a^x \log a\dfrac{d}{dx}(a^x) = a^x \log a
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Take log (x > 0 so it is allowed).लघुगणक लीजिए (x > 0, इसलिए अनुमत)।Log lijiye (x > 0, isliye allowed).y = x^{\sin x} \Rightarrow \log y = \sin x \log x
- Product rule on the right.दाएँ पक्ष पर गुणनफल नियम।Right side par product rule.\dfrac{1}{y}\dfrac{dy}{dx} = \sin x \cdot \dfrac{1}{x} + \log x \cdot \cos x
- Multiply by y = x^{\sin x}.y = x^{\sin x} से गुणा।y = x^{\sin x} se multiply.\dfrac{dy}{dx} = x^{\sin x}\left[\dfrac{\sin x}{x} + \cos x \log x\right]
✔ \dfrac{dy}{dx} = x^{\sin x - 1} \cdot \sin x + x^{\sin x} \cdot \cos x \log x\dfrac{dy}{dx} = x^{\sin x - 1} \cdot \sin x + x^{\sin x} \cdot \cos x \log x\dfrac{dy}{dx} = x^{\sin x - 1} \cdot \sin x + x^{\sin x} \cdot \cos x \log x
Solution — step by step (5)हल — चरण दर चरण (5)Hal — step by step (5)
- Split into three parts; a^b is a constant. (Never take log of a sum!)तीन भागों में बाँटिए; a^b अचर है। (योग का लघुगणक कभी नहीं!)Teen hisson mein baantiye; a^b constant hai. (Sum ka log kabhi nahi!)u = y^x,\ v = x^y,\ w = x^x:\ \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx} = 0 \quad (1)
- Log-differentiate u (product rule, chain rule on log y).u का लघुगणकीय अवकलन (गुणनफल नियम, log y पर श्रृंखला नियम)।u ka logarithmic differentiation (product rule, log y par chain rule).\log u = x \log y \Rightarrow \dfrac{du}{dx} = y^x\left(\dfrac{x}{y}\dfrac{dy}{dx} + \log y\right) \quad (2)
- Same for v.v के लिए वही।v ke liye wahi.\log v = y \log x \Rightarrow \dfrac{dv}{dx} = x^y\left(\dfrac{y}{x} + \log x \dfrac{dy}{dx}\right) \quad (3)
- Same for w.w के लिए वही।w ke liye wahi.\log w = x \log x \Rightarrow \dfrac{dw}{dx} = x^x(1 + \log x) \quad (4)
- Put (2), (3), (4) in (1) and collect the \dfrac{dy}{dx} terms.(2), (3), (4) को (1) में रखकर \dfrac{dy}{dx} वाले पद एक ओर कीजिए।(2), (3), (4) ko (1) mein rakhkar \dfrac{dy}{dx} wale terms ek taraf kijiye.\dfrac{dy}{dx}\left(x \cdot y^{x - 1} + x^y \log x\right) = -x^x(1 + \log x) - y \cdot x^{y - 1} - y^x \log y
✔ \dfrac{dy}{dx} = \dfrac{-\left[y^x \log y + y \cdot x^{y - 1} + x^x(1 + \log x)\right]}{x \cdot y^{x - 1} + x^y \log x}\dfrac{dy}{dx} = \dfrac{-\left[y^x \log y + y \cdot x^{y - 1} + x^x(1 + \log x)\right]}{x \cdot y^{x - 1} + x^y \log x}\dfrac{dy}{dx} = \dfrac{-\left[y^x \log y + y \cdot x^{y - 1} + x^x(1 + \log x)\right]}{x \cdot y^{x - 1} + x^y \log x}
(Exercise 5.5, Q1) Differentiate \cos x \cdot \cos 2x \cdot \cos 3x w.r.t. x.(प्रश्नावली 5.5, प्र. 1) \cos x \cdot \cos 2x \cdot \cos 3x का x के सापेक्ष अवकलन कीजिए।(Exercise 5.5, Q1) \cos x \cdot \cos 2x \cdot \cos 3x ko x ke respect mein differentiate kijiye.
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-\cos x \cos 2x \cos 3x\,[\tan x + 2\tan 2x + 3\tan 3x]-\cos x \cos 2x \cos 3x\,[\tan x + 2\tan 2x + 3\tan 3x]-\cos x \cos 2x \cos 3x\,[\tan x + 2\tan 2x + 3\tan 3x]
(Exercise 5.5, Q4) Differentiate x^x - 2^{\sin x} w.r.t. x.(प्रश्नावली 5.5, प्र. 4) x^x - 2^{\sin x} का x के सापेक्ष अवकलन कीजिए।(Exercise 5.5, Q4) x^x - 2^{\sin x} ko x ke respect mein differentiate kijiye.
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x^x(1 + \log x) - 2^{\sin x} \cos x \log 2x^x(1 + \log x) - 2^{\sin x} \cos x \log 2x^x(1 + \log x) - 2^{\sin x} \cos x \log 2
(Exercise 5.5, Q13) Find \dfrac{dy}{dx} if y^x = x^y.(प्रश्नावली 5.5, प्र. 13) यदि y^x = x^y हो, तो \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.5, Q13) Agar y^x = x^y ho, to \dfrac{dy}{dx} nikaaliye.
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\dfrac{y}{x}\left(\dfrac{y - x \log y}{x - y \log x}\right)\dfrac{y}{x}\left(\dfrac{y - x \log y}{x - y \log x}\right)\dfrac{y}{x}\left(\dfrac{y - x \log y}{x - y \log x}\right)
(Exercise 5.5, Q15) Find \dfrac{dy}{dx} if xy = e^{(x - y)}.(प्रश्नावली 5.5, प्र. 15) यदि xy = e^{(x - y)} हो, तो \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.5, Q15) Agar xy = e^{(x - y)} ho, to \dfrac{dy}{dx} nikaaliye.
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\dfrac{y(x - 1)}{x(y + 1)}\dfrac{y(x - 1)}{x(y + 1)}\dfrac{y(x - 1)}{x(y + 1)}
(Exercise 5.5, Q16) Find the derivative of f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) and hence find f'(1).(प्रश्नावली 5.5, प्र. 16) f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) का अवकलज ज्ञात कीजिए और f'(1) निकालिए।(Exercise 5.5, Q16) f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8) ka derivative nikaaliye aur f'(1) nikaaliye.
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f'(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]; f'(1) = 120f'(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]; f'(1) = 120f'(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]; f'(1) = 120
5.6Derivatives of Functions in Parametric Formsप्राचलिक रूपों में फलनों के अवकलजParametric form mein functions ke derivatives
Sometimes the relation between x and y is neither explicit nor implicit, but each of them is linked separately to a third variable. A relation x = f(t), y = g(t) is said to be in parametric form with t as a parameter. By the chain rule \dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}, so \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \left(\text{provided } f'(t) \ne 0\right)कभी-कभी x और y के बीच संबंध न स्पष्ट होता है न अस्पष्ट, बल्कि दोनों अलग-अलग किसी तीसरे चर से जुड़े होते हैं। संबंध x = f(t), y = g(t) को t प्राचल वाला प्राचलिक रूप कहते हैं। श्रृंखला नियम से \dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}, अतः \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \left(\text{यदि } f'(t) \ne 0\right)Kabhi-kabhi x aur y ka relation na explicit hota hai na implicit, balki dono alag-alag kisi teesre variable se jude hote hain. Relation x = f(t), y = g(t) ko t parameter wala parametric form kehte hain. Chain rule se \dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}, isliye \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \left(\text{agar } f'(t) \ne 0\right)x = f(t), y = g(t) ⇒ dy/dx = g′(t)/f′(t), when f′(t) ≠ 0.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Differentiate each w.r.t. the parameter θ.हर एक का प्राचल θ के सापेक्ष अवकलन।Har ek ko parameter θ ke respect mein differentiate kijiye.\dfrac{dx}{d\theta} = -a\sin\theta,\quad \dfrac{dy}{d\theta} = a\cos\theta
- Divide.भाग दीजिए।Divide kijiye.\dfrac{dy}{dx} = \dfrac{a\cos\theta}{-a\sin\theta}
✔ \dfrac{dy}{dx} = -\cot\theta\dfrac{dy}{dx} = -\cot\theta\dfrac{dy}{dx} = -\cot\theta
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Differentiate w.r.t. t.t के सापेक्ष अवकलन।t ke respect mein differentiate.\dfrac{dx}{dt} = 2at,\quad \dfrac{dy}{dt} = 2a
- Divide.भाग दीजिए।Divide kijiye.\dfrac{dy}{dx} = \dfrac{2a}{2at}
✔ \dfrac{dy}{dx} = \dfrac{1}{t}\dfrac{dy}{dx} = \dfrac{1}{t}\dfrac{dy}{dx} = \dfrac{1}{t}
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- Differentiate w.r.t. θ.θ के सापेक्ष अवकलन।θ ke respect mein differentiate.\dfrac{dx}{d\theta} = a(1 + \cos\theta),\quad \dfrac{dy}{d\theta} = a\sin\theta
- Half-angle formulas: \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, 1 + \cos\theta = 2\cos^2\frac{\theta}{2}.अर्ध-कोण सूत्र: \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, 1 + \cos\theta = 2\cos^2\frac{\theta}{2}।Half-angle formulas: \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, 1 + \cos\theta = 2\cos^2\frac{\theta}{2}.\dfrac{dy}{dx} = \dfrac{a\sin\theta}{a(1 + \cos\theta)} = \dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}
✔ \dfrac{dy}{dx} = \tan\dfrac{\theta}{2}. Note: \dfrac{dy}{dx} is expressed in terms of the parameter only, without directly involving x and y.\dfrac{dy}{dx} = \tan\dfrac{\theta}{2}। ध्यान दीजिए: \dfrac{dy}{dx} केवल प्राचल के पदों में है, x और y सीधे नहीं आते।\dfrac{dy}{dx} = \tan\dfrac{\theta}{2}. Dhyaan dijiye: \dfrac{dy}{dx} sirf parameter mein hai, x aur y seedhe nahi aate.
A parametric curve: x = a(θ + sin θ), y = a(1 − cos θ)प्राचलिक वक्र: x = a(θ + sin θ), y = a(1 − cos θ)Parametric curve: x = a(θ + sin θ), y = a(1 − cos θ)NCERT Example 33 · 8 tappable partsOpenSolution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Choose a parametric form of the curve.वक्र का प्राचलिक रूप चुनिए।Curve ka parametric form chuniye.x = a\cos^3\theta,\ y = a\sin^3\theta
- Check: it satisfies the equation.जाँच: यह समीकरण को संतुष्ट करता है।Check: ye equation ko satisfy karta hai.x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}(\cos^2\theta + \sin^2\theta) = a^{\frac{2}{3}}
- Chain rule on \cos^3\theta and \sin^3\theta.\cos^3\theta और \sin^3\theta पर श्रृंखला नियम।\cos^3\theta aur \sin^3\theta par chain rule.\dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta,\quad \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta
- Divide and cancel.भाग देकर काटिए।Divide karke cancel kijiye.\dfrac{dy}{dx} = \dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\tan\theta
✔ \dfrac{dy}{dx} = -\tan\theta = -\sqrt[3]{\dfrac{y}{x}} (since \dfrac{y}{x} = \tan^3\theta).\dfrac{dy}{dx} = -\tan\theta = -\sqrt[3]{\dfrac{y}{x}} (क्योंकि \dfrac{y}{x} = \tan^3\theta)।\dfrac{dy}{dx} = -\tan\theta = -\sqrt[3]{\dfrac{y}{x}} (kyunki \dfrac{y}{x} = \tan^3\theta).
(Exercise 5.6, Q1) x = 2at^2,\ y = at^4. Find \dfrac{dy}{dx}.(प्रश्नावली 5.6, प्र. 1) x = 2at^2,\ y = at^4। \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.6, Q1) x = 2at^2,\ y = at^4. \dfrac{dy}{dx} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
t^2t^2t^2
(Exercise 5.6, Q3) x = \sin t,\ y = \cos 2t. Find \dfrac{dy}{dx}.(प्रश्नावली 5.6, प्र. 3) x = \sin t,\ y = \cos 2t। \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.6, Q3) x = \sin t,\ y = \cos 2t. \dfrac{dy}{dx} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
-4\sin t-4\sin t-4\sin t
(Exercise 5.6, Q4) x = 4t,\ y = \dfrac{4}{t}. Find \dfrac{dy}{dx}.(प्रश्नावली 5.6, प्र. 4) x = 4t,\ y = \dfrac{4}{t}। \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.6, Q4) x = 4t,\ y = \dfrac{4}{t}. \dfrac{dy}{dx} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
-\dfrac{1}{t^2}-\dfrac{1}{t^2}-\dfrac{1}{t^2}
(Exercise 5.6, Q6) x = a(\theta - \sin\theta),\ y = a(1 + \cos\theta). Find \dfrac{dy}{dx}.(प्रश्नावली 5.6, प्र. 6) x = a(\theta - \sin\theta),\ y = a(1 + \cos\theta)। \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.6, Q6) x = a(\theta - \sin\theta),\ y = a(1 + \cos\theta). \dfrac{dy}{dx} nikaaliye.
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-\cot\dfrac{\theta}{2}-\cot\dfrac{\theta}{2}-\cot\dfrac{\theta}{2}
(Exercise 5.6, Q9) x = a\sec\theta,\ y = b\tan\theta. Find \dfrac{dy}{dx}.(प्रश्नावली 5.6, प्र. 9) x = a\sec\theta,\ y = b\tan\theta। \dfrac{dy}{dx} ज्ञात कीजिए।(Exercise 5.6, Q9) x = a\sec\theta,\ y = b\tan\theta. \dfrac{dy}{dx} nikaaliye.
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\dfrac{b}{a}\,\text{cosec}\,\theta\dfrac{b}{a}\,\text{cosec}\,\theta\dfrac{b}{a}\,\text{cosec}\,\theta
5.7Second Order Derivativeद्वितीय कोटि का अवकलजSecond Order Derivative
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- First derivative.प्रथम अवकलज।First derivative.\dfrac{dy}{dx} = 3x^2 + \sec^2 x
- Differentiate again; chain rule on (\sec x)^2.फिर से अवकलन; (\sec x)^2 पर श्रृंखला नियम।Dobara differentiate; (\sec x)^2 par chain rule.\dfrac{d^2y}{dx^2} = 6x + 2\sec x \cdot \sec x \tan x
✔ \dfrac{d^2y}{dx^2} = 6x + 2\sec^2 x \tan x\dfrac{d^2y}{dx^2} = 6x + 2\sec^2 x \tan x\dfrac{d^2y}{dx^2} = 6x + 2\sec^2 x \tan x
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- First derivative.प्रथम अवकलज।First derivative.\dfrac{dy}{dx} = A\cos x - B\sin x
- Second derivative is exactly -y.द्वितीय अवकलज ठीक -y है।Second derivative theek -y hai.\dfrac{d^2y}{dx^2} = -A\sin x - B\cos x = -y
✔ \dfrac{d^2y}{dx^2} + y = 0. Hence proved.\dfrac{d^2y}{dx^2} + y = 0। अतः सिद्ध हुआ।\dfrac{d^2y}{dx^2} + y = 0. Isliye prove hua.
Solution — step by step (4)हल — चरण दर चरण (4)Hal — step by step (4)
- Chain rule on each term.हर पद पर श्रृंखला नियम।Har term par chain rule.\dfrac{dy}{dx} = 6e^{2x} + 6e^{3x} = 6(e^{2x} + e^{3x})
- Differentiate again.फिर से अवकलन।Dobara differentiate.\dfrac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x} = 6(2e^{2x} + 3e^{3x})
- Substitute into the left side.बाएँ पक्ष में रखिए।Left side mein rakhiye.6(2e^{2x} + 3e^{3x}) - 30(e^{2x} + e^{3x}) + 6(3e^{2x} + 2e^{3x})
- Collect the terms.पदों को इकट्ठा कीजिए।Terms ko ikattha kijiye.= (12 - 30 + 18)e^{2x} + (18 - 30 + 12)e^{3x} = 0
✔ \dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0. Hence proved.\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0। अतः सिद्ध हुआ।\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0. Isliye prove hua.
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- Clear the fraction first — it makes the next step easy.पहले भिन्न हटाइए — अगला कदम आसान हो जाता है।Pehle fraction hataiye — agla step aasaan ho jaata hai.\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} \Rightarrow \sqrt{1 - x^2}\,\dfrac{dy}{dx} = 1
- Differentiate both sides (product rule; derivative of 1 is 0).दोनों पक्षों का अवकलन (गुणनफल नियम; 1 का अवकलज 0)।Dono sides differentiate (product rule; 1 ka derivative 0).\sqrt{1 - x^2}\,\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} \cdot \dfrac{-2x}{2\sqrt{1 - x^2}} = 0
- Multiply by \sqrt{1 - x^2}. (Alternatively: square first, (1 - x^2)y_1^2 = 1, then differentiate: (1 - x^2) \cdot 2y_1y_2 + y_1^2(-2x) = 0.)\sqrt{1 - x^2} से गुणा। (वैकल्पिक: पहले वर्ग कीजिए, (1 - x^2)y_1^2 = 1, फिर अवकलन: (1 - x^2) \cdot 2y_1y_2 + y_1^2(-2x) = 0।)\sqrt{1 - x^2} se multiply. (Doosra tareeka: pehle square kijiye, (1 - x^2)y_1^2 = 1, phir differentiate: (1 - x^2) \cdot 2y_1y_2 + y_1^2(-2x) = 0.)(1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 0
✔ (1 - x^2)y_2 - xy_1 = 0. Hence shown.(1 - x^2)y_2 - xy_1 = 0। अतः दर्शाया गया।(1 - x^2)y_2 - xy_1 = 0. Isliye dikhaya gaya.
(Exercise 5.7, Q2) Find the second order derivative of x^{20}.(प्रश्नावली 5.7, प्र. 2) x^{20} का द्वितीय कोटि का अवकलज ज्ञात कीजिए।(Exercise 5.7, Q2) x^{20} ka second order derivative nikaaliye.
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380x^{18}380x^{18}380x^{18}
(Exercise 5.7, Q3) Find the second order derivative of x \cos x.(प्रश्नावली 5.7, प्र. 3) x \cos x का द्वितीय कोटि का अवकलज ज्ञात कीजिए।(Exercise 5.7, Q3) x \cos x ka second order derivative nikaaliye.
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-x\cos x - 2\sin x-x\cos x - 2\sin x-x\cos x - 2\sin x
(Exercise 5.7, Q5) Find the second order derivative of x^3 \log x.(प्रश्नावली 5.7, प्र. 5) x^3 \log x का द्वितीय कोटि का अवकलज ज्ञात कीजिए।(Exercise 5.7, Q5) x^3 \log x ka second order derivative nikaaliye.
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x(5 + 6\log x)x(5 + 6\log x)x(5 + 6\log x)
(Exercise 5.7, Q6) Find the second order derivative of e^x \sin 5x.(प्रश्नावली 5.7, प्र. 6) e^x \sin 5x का द्वितीय कोटि का अवकलज ज्ञात कीजिए।(Exercise 5.7, Q6) e^x \sin 5x ka second order derivative nikaaliye.
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2e^x(5\cos 5x - 12\sin 5x)2e^x(5\cos 5x - 12\sin 5x)2e^x(5\cos 5x - 12\sin 5x)
(Exercise 5.7, Q8) Find the second order derivative of \tan^{-1} x.(प्रश्नावली 5.7, प्र. 8) \tan^{-1} x का द्वितीय कोटि का अवकलज ज्ञात कीजिए।(Exercise 5.7, Q8) \tan^{-1} x ka second order derivative nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
-\dfrac{2x}{(1 + x^2)^2}-\dfrac{2x}{(1 + x^2)^2}-\dfrac{2x}{(1 + x^2)^2}
(Exercise 5.7, Q12) If y = \cos^{-1} x, find \dfrac{d^2y}{dx^2} in terms of y alone.(प्रश्नावली 5.7, प्र. 12) यदि y = \cos^{-1} x, तो \dfrac{d^2y}{dx^2} केवल y के पदों में ज्ञात कीजिए।(Exercise 5.7, Q12) Agar y = \cos^{-1} x, to \dfrac{d^2y}{dx^2} sirf y mein nikaaliye.
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-\cot y\,\text{cosec}^2\,y-\cot y\,\text{cosec}^2\,y-\cot y\,\text{cosec}^2\,y
Miscellaneous Examplesविविध उदाहरणMiscellaneous Examples
Solution — step by step (5)हल — चरण दर चरण (5)Hal — step by step (5)
- Write roots as powers; defined for x > -\frac{2}{3}.मूलों को घात में लिखिए; x > -\frac{2}{3} के लिए परिभाषित।Roots ko power mein likhiye; x > -\frac{2}{3} ke liye defined.\text{(i)}\ y = (3x + 2)^{\frac{1}{2}} + (2x^2 + 4)^{-\frac{1}{2}},\quad x > -\dfrac{2}{3}
- Chain rule on each term.हर पद पर श्रृंखला नियम।Har term par chain rule.\dfrac{dy}{dx} = \dfrac{1}{2}(3x + 2)^{-\frac{1}{2}} \cdot 3 + \left(-\dfrac{1}{2}\right)(2x^2 + 4)^{-\frac{3}{2}} \cdot 4x
- Simplify (valid for x > -\frac{2}{3}).सरल कीजिए (x > -\frac{2}{3} के लिए)।Simplify kijiye (x > -\frac{2}{3} ke liye).\text{(i)}\ \dfrac{dy}{dx} = \dfrac{3}{2\sqrt{3x + 2}} - \dfrac{2x}{(2x^2 + 4)^{\frac{3}{2}}}
- Change of base; \log x > 0 needs x > 1.आधार परिवर्तन; \log x > 0 के लिए x > 1 चाहिए।Change of base; \log x > 0 ke liye x > 1 chahiye.\text{(ii)}\ y = \log_7(\log x) = \dfrac{\log(\log x)}{\log 7},\quad x > 1
- Chain rule.श्रृंखला नियम।Chain rule.\dfrac{dy}{dx} = \dfrac{1}{\log 7} \cdot \dfrac{1}{\log x} \cdot \dfrac{1}{x}
✔ (i) \dfrac{3}{2\sqrt{3x + 2}} - \dfrac{2x}{(2x^2 + 4)^{\frac{3}{2}}} (ii) \dfrac{1}{x \log 7 \log x}(i) \dfrac{3}{2\sqrt{3x + 2}} - \dfrac{2x}{(2x^2 + 4)^{\frac{3}{2}}} (ii) \dfrac{1}{x \log 7 \log x}(i) \dfrac{3}{2\sqrt{3x + 2}} - \dfrac{2x}{(2x^2 + 4)^{\frac{3}{2}}} (ii) \dfrac{1}{x \log 7 \log x}
Solution — step by step (6)हल — चरण दर चरण (6)Hal — step by step (6)
- Rewrite sin as cos. (This last equality needs \frac{\pi}{2} - x \in [0, \pi], i.e., -\frac{\pi}{2} \le x \le \frac{\pi}{2} — NCERT's answer -1 is for these x.)sin को cos में लिखिए। (यह अंतिम समता \frac{\pi}{2} - x \in [0, \pi], अर्थात् -\frac{\pi}{2} \le x \le \frac{\pi}{2} के लिए है — NCERT का उत्तर -1 इन्हीं x के लिए है।)sin ko cos mein likhiye. (Ye aakhri equality \frac{\pi}{2} - x \in [0, \pi], yaani -\frac{\pi}{2} \le x \le \frac{\pi}{2} ke liye hai — NCERT ka answer -1 inhi x ke liye hai.)\text{(i)}\ \cos^{-1}(\sin x) = \cos^{-1}\left(\cos\left(\dfrac{\pi}{2} - x\right)\right) = \dfrac{\pi}{2} - x \Rightarrow f'(x) = -1
- Defined where \cos x \ne -1, i.e., x not an odd multiple of π; cancel \cos\frac{x}{2} \ne 0.जहाँ \cos x \ne -1, अर्थात् x, π का विषम गुणज नहीं; \cos\frac{x}{2} \ne 0 को काटिए।Jahan \cos x \ne -1, yaani x, π ka odd multiple nahi; \cos\frac{x}{2} \ne 0 ko cancel kijiye.\text{(ii)}\ \dfrac{\sin x}{1 + \cos x} = \dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan\dfrac{x}{2}
- Inverse undoes tan.प्रतिलोम tan को निरस्त करता है।Inverse tan ko undo karta hai.\text{(ii)}\ f(x) = \tan^{-1}\left(\tan\dfrac{x}{2}\right) = \dfrac{x}{2} \Rightarrow f'(x) = \dfrac{1}{2}
- The expression is always positive and ≤ 1, so f is defined for every real x. Substitute.व्यंजक सदा धनात्मक और ≤ 1 है, अतः f हर वास्तविक x पर परिभाषित है। प्रतिस्थापन कीजिए।Expression hamesha positive aur ≤ 1 hai, isliye f har real x par defined hai. Substitute kijiye.\text{(iii)}\ 2^x = \tan\theta:\ \dfrac{2^{x + 1}}{1 + 4^x} = \dfrac{2\tan\theta}{1 + \tan^2\theta} = \sin 2\theta
- Valid when 2\theta \le \frac{\pi}{2}, i.e., 2^x \le 1, i.e., x \le 0.मान्य जब 2\theta \le \frac{\pi}{2}, अर्थात् 2^x \le 1, अर्थात् x \le 0।Sahi jab 2\theta \le \frac{\pi}{2}, yaani 2^x \le 1, yaani x \le 0.f(x) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}(2^x)
- Chain rule with \dfrac{d}{dx}2^x = 2^x \log 2.श्रृंखला नियम, \dfrac{d}{dx}2^x = 2^x \log 2।Chain rule, \dfrac{d}{dx}2^x = 2^x \log 2.f'(x) = 2 \cdot \dfrac{1}{1 + (2^x)^2} \cdot 2^x \log 2 = \dfrac{2^{x + 1}\log 2}{1 + 4^x}
✔ (i) -1 (ii) \dfrac{1}{2} (iii) \dfrac{2^{x + 1}\log 2}{1 + 4^x} — NCERT's answers. Topper note on (iii): for x > 0, 2\theta > \frac{\pi}{2} and \sin^{-1}(\sin 2\theta) = \pi - 2\theta, so there the derivative is -\dfrac{2^{x + 1}\log 2}{1 + 4^x} (f is not differentiable at x = 0).(i) -1 (ii) \dfrac{1}{2} (iii) \dfrac{2^{x + 1}\log 2}{1 + 4^x} — NCERT के उत्तर। (iii) पर टॉपर टिप्पणी: x > 0 के लिए 2\theta > \frac{\pi}{2} और \sin^{-1}(\sin 2\theta) = \pi - 2\theta, इसलिए वहाँ अवकलज -\dfrac{2^{x + 1}\log 2}{1 + 4^x} है (x = 0 पर f अवकलनीय नहीं)।(i) -1 (ii) \dfrac{1}{2} (iii) \dfrac{2^{x + 1}\log 2}{1 + 4^x} — NCERT ke answers. (iii) par topper note: x > 0 ke liye 2\theta > \frac{\pi}{2} aur \sin^{-1}(\sin 2\theta) = \pi - 2\theta, isliye wahan derivative -\dfrac{2^{x + 1}\log 2}{1 + 4^x} hai (x = 0 par f differentiable nahi).
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- For 0 < x < \pi, \sin x > 0, so log is allowed.0 < x < \pi के लिए \sin x > 0, इसलिए लघुगणक ले सकते हैं।0 < x < \pi ke liye \sin x > 0, isliye log le sakte hain.y = (\sin x)^{\sin x} \Rightarrow \log y = \sin x \log(\sin x)
- Product rule + chain rule.गुणनफल नियम + श्रृंखला नियम।Product rule + chain rule.\dfrac{1}{y}\dfrac{dy}{dx} = \cos x \log(\sin x) + \sin x \cdot \dfrac{1}{\sin x} \cdot \cos x
- Simplify.सरल कीजिए।Simplify kijiye.\dfrac{1}{y}\dfrac{dy}{dx} = (1 + \log(\sin x))\cos x
✔ \dfrac{dy}{dx} = (1 + \log(\sin x))(\sin x)^{\sin x}\cos x\dfrac{dy}{dx} = (1 + \log(\sin x))(\sin x)^{\sin x}\cos x\dfrac{dy}{dx} = (1 + \log(\sin x))(\sin x)^{\sin x}\cos x
Solution — step by step (3)हल — चरण दर चरण (3)Hal — step by step (3)
- \dfrac{d}{du}a^u = a^u \log a with u = t + \frac{1}{t} (t ≠ 0).\dfrac{d}{du}a^u = a^u \log a, जहाँ u = t + \frac{1}{t} (t ≠ 0)।\dfrac{d}{du}a^u = a^u \log a, jahan u = t + \frac{1}{t} (t ≠ 0).\dfrac{dy}{dt} = a^{t + \frac{1}{t}}\left(1 - \dfrac{1}{t^2}\right)\log a
- Power rule + chain rule; \dfrac{dx}{dt} \ne 0 only if t \ne \pm 1.घात नियम + श्रृंखला नियम; \dfrac{dx}{dt} \ne 0 केवल यदि t \ne \pm 1।Power rule + chain rule; \dfrac{dx}{dt} \ne 0 sirf agar t \ne \pm 1.\dfrac{dx}{dt} = a\left(t + \dfrac{1}{t}\right)^{a - 1}\left(1 - \dfrac{1}{t^2}\right)
- Divide (t ≠ ±1) and cancel \left(1 - \frac{1}{t^2}\right).भाग दीजिए (t ≠ ±1) और \left(1 - \frac{1}{t^2}\right) काटिए।Divide kijiye (t ≠ ±1) aur \left(1 - \frac{1}{t^2}\right) cancel kijiye.\dfrac{dy}{dx} = \dfrac{a^{t + \frac{1}{t}}\left(1 - \frac{1}{t^2}\right)\log a}{a\left(t + \frac{1}{t}\right)^{a - 1}\left(1 - \frac{1}{t^2}\right)}
✔ \dfrac{dy}{dx} = \dfrac{a^{t + \frac{1}{t}}\log a}{a\left(t + \frac{1}{t}\right)^{a - 1}},\quad t \ne 0, \pm 1\dfrac{dy}{dx} = \dfrac{a^{t + \frac{1}{t}}\log a}{a\left(t + \frac{1}{t}\right)^{a - 1}},\quad t \ne 0, \pm 1\dfrac{dy}{dx} = \dfrac{a^{t + \frac{1}{t}}\log a}{a\left(t + \frac{1}{t}\right)^{a - 1}},\quad t \ne 0, \pm 1
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
- "w.r.t. another function" = treat x as the parameter."किसी अन्य फलन के सापेक्ष" = x को प्राचल मानिए।"Kisi aur function ke respect mein" = x ko parameter maaniye.u = \sin^2 x,\ v = e^{\cos x}:\ \dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}
- Chain rule on each.हर एक पर श्रृंखला नियम।Har ek par chain rule.\dfrac{du}{dx} = 2\sin x\cos x,\quad \dfrac{dv}{dx} = e^{\cos x}(-\sin x)
✔ \dfrac{du}{dv} = \dfrac{2\sin x\cos x}{-\sin x\, e^{\cos x}} = -\dfrac{2\cos x}{e^{\cos x}}\dfrac{du}{dv} = \dfrac{2\sin x\cos x}{-\sin x\, e^{\cos x}} = -\dfrac{2\cos x}{e^{\cos x}}\dfrac{du}{dv} = \dfrac{2\sin x\cos x}{-\sin x\, e^{\cos x}} = -\dfrac{2\cos x}{e^{\cos x}}
(Miscellaneous Exercise, Q1) Differentiate (3x^2 - 9x + 5)^9 w.r.t. x.(विविध प्रश्नावली, प्र. 1) (3x^2 - 9x + 5)^9 का x के सापेक्ष अवकलन कीजिए।(Miscellaneous Exercise, Q1) (3x^2 - 9x + 5)^9 ko x ke respect mein differentiate kijiye.
Show answerउत्तर देखिएAnswer dekhiye
27(3x^2 - 9x + 5)^8(2x - 3)27(3x^2 - 9x + 5)^8(2x - 3)27(3x^2 - 9x + 5)^8(2x - 3)
(Miscellaneous Exercise, Q10) Differentiate x^x + x^a + a^x + a^a w.r.t. x, for some fixed a > 0 and x > 0.(विविध प्रश्नावली, प्र. 10) x^x + x^a + a^x + a^a का x के सापेक्ष अवकलन कीजिए, जहाँ a > 0 नियत और x > 0।(Miscellaneous Exercise, Q10) x^x + x^a + a^x + a^a ko x ke respect mein differentiate kijiye, jahan a > 0 fixed aur x > 0.
Show answerउत्तर देखिएAnswer dekhiye
x^x(1 + \log x) + ax^{a - 1} + a^x \log ax^x(1 + \log x) + ax^{a - 1} + a^x \log ax^x(1 + \log x) + ax^{a - 1} + a^x \log a
(Miscellaneous Exercise, Q12) Find \dfrac{dy}{dx} if y = 12(1 - \cos t),\ x = 10(t - \sin t),\ -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}.(विविध प्रश्नावली, प्र. 12) यदि y = 12(1 - \cos t),\ x = 10(t - \sin t),\ -\dfrac{\pi}{2} < t < \dfrac{\pi}{2} है, तो \dfrac{dy}{dx} ज्ञात कीजिए।(Miscellaneous Exercise, Q12) Agar y = 12(1 - \cos t),\ x = 10(t - \sin t),\ -\dfrac{\pi}{2} < t < \dfrac{\pi}{2} hai, to \dfrac{dy}{dx} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
\dfrac{6}{5}\cot\dfrac{t}{2}\dfrac{6}{5}\cot\dfrac{t}{2}\dfrac{6}{5}\cot\dfrac{t}{2}
(Miscellaneous Exercise, Q13) Find \dfrac{dy}{dx} if y = \sin^{-1} x + \sin^{-1}\sqrt{1 - x^2},\ 0 < x < 1.(विविध प्रश्नावली, प्र. 13) यदि y = \sin^{-1} x + \sin^{-1}\sqrt{1 - x^2},\ 0 < x < 1 है, तो \dfrac{dy}{dx} ज्ञात कीजिए।(Miscellaneous Exercise, Q13) Agar y = \sin^{-1} x + \sin^{-1}\sqrt{1 - x^2},\ 0 < x < 1 hai, to \dfrac{dy}{dx} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
0 (here y = \dfrac{\pi}{2}, a constant)0 (यहाँ y = \dfrac{\pi}{2}, एक अचर)0 (yahan y = \dfrac{\pi}{2}, ek constant)
(Miscellaneous Exercise, Q17) If x = a(\cos t + t\sin t) and y = a(\sin t - t\cos t), find \dfrac{d^2y}{dx^2}.(विविध प्रश्नावली, प्र. 17) यदि x = a(\cos t + t\sin t) और y = a(\sin t - t\cos t) है, तो \dfrac{d^2y}{dx^2} ज्ञात कीजिए।(Miscellaneous Exercise, Q17) Agar x = a(\cos t + t\sin t) aur y = a(\sin t - t\cos t) hai, to \dfrac{d^2y}{dx^2} nikaaliye.
Show answerउत्तर देखिएAnswer dekhiye
\dfrac{\sec^3 t}{at},\ 0 < t < \dfrac{\pi}{2}\dfrac{\sec^3 t}{at},\ 0 < t < \dfrac{\pi}{2}\dfrac{\sec^3 t}{at},\ 0 < t < \dfrac{\pi}{2}
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