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NCERT Solutions · Class 12 · Chapter 5: Continuity and Differentiability
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5
Exercise 5.5: Logarithmic differentiation. For products, quotients and powers like \(u^v\) (variable base and exponent), take \(\log\) of both sides first: \(\log y = v\log u\), then \(\tfrac1y\tfrac{dy}{dx} = v'\log u + \tfrac{vu'}{u}\). For a sum such as \(x^x + (\sin x)^x\), differentiate each term separately (logs do not split sums).
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Exercise 5.5 questions and solutions
Exercise 5.5, Question 1
\(\cos x \cdot \cos 2x \cdot \cos 3x\)
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\[\log y = \log\cos x + \log\cos 2x + \log\cos 3x\] (where the cosines are positive; the result holds generally). \[\dfrac{y'}{y} = -\tan x - 2\tan 2x - 3\tan 3x\] Answer: \[-\cos x\cos 2x\cos 3x\,(\tan x + 2\tan 2x + 3\tan 3x)\]
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Exercise 5.5, Question 2
\(\sqrt{\dfrac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)}}\)
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\[\log y = \tfrac12[\log(x - 1) + \log(x - 2) - \log(x - 3) - \log(x - 4) - \log(x - 5)]\] (for \(x > 5\)). Differentiate and multiply by y. Answer: \[\dfrac12\sqrt{\dfrac{(x - 1)(x - 2)}{(x - 3)(x - 4)(x - 5)}}\left[\dfrac{1}{x - 1} + \dfrac{1}{x - 2} - \dfrac{1}{x - 3} - \dfrac{1}{x - 4} - \dfrac{1}{x - 5}\right]\]
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Exercise 5.5, Question 3
\((\log x)^{\cos x}\)
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\(\log y = \cos x\log(\log x)\) (\(x > 1\)). \[\dfrac{y'}{y} = -\sin x\log(\log x) + \dfrac{\cos x}{x\log x}\] Answer: \[(\log x)^{\cos x}\left[\dfrac{\cos x}{x\log x} - \sin x\log(\log x)\right]\]
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Exercise 5.5, Question 4
\(x^x - 2^{\sin x}\)
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\(u = x^x\): \(\log u = x\log x\), \(u' = x^x(1 + \log x)\). \[(2^{\sin x})' = 2^{\sin x}\log 2 \cdot \cos x\] Answer: \(x^x(1 + \log x) - 2^{\sin x}\cos x\log 2\)
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Exercise 5.5, Question 5
\((x + 3)^2(x + 4)^3(x + 5)^4\)
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\[\log y = 2\log(x + 3) + 3\log(x + 4) + 4\log(x + 5)\] \[\begin{aligned}\dfrac{y'}{y} &= \dfrac{2}{x + 3} + \dfrac{3}{x + 4} + \dfrac{4}{x + 5} \\ &= \dfrac{9x^2 + 70x + 133}{(x + 3)(x + 4)(x + 5)}\end{aligned}\] Answer: \[(x + 3)(x + 4)^2(x + 5)^3(9x^2 + 70x + 133)\]
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Exercise 5.5, Question 6
\(\left(x + \dfrac1x\right)^x + x^{1 + \frac1x}\)
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\(u = \left(x + \tfrac1x\right)^x\): \(\log u = x\log\left(x + \tfrac1x\right)\), \[\dfrac{u'}{u} = \log\left(x + \tfrac1x\right) + \dfrac{x^2 - 1}{x^2 + 1}\] \(v = x^{1 + \frac1x}\): \(\log v = \left(1 + \tfrac1x\right)\log x\), \[\begin{aligned}\dfrac{v'}{v} &= -\dfrac{\log x}{x^2} + \dfrac{x + 1}{x^2} \\ &= \dfrac{x + 1 - \log x}{x^2}\end{aligned}\] Answer: \[\left(x + \dfrac1x\right)^x\left[\dfrac{x^2 - 1}{x^2 + 1} + \log\left(x + \dfrac1x\right)\right] + x^{1 + \frac1x}\cdot\dfrac{x + 1 - \log x}{x^2}\]
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Exercise 5.5, Question 7
\((\log x)^x + x^{\log x}\)
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\(u = (\log x)^x\): \[\dfrac{u'}{u} = \log(\log x) + \dfrac{1}{\log x}\] (\(x > 1\)). \(v = x^{\log x}\): \(\log v = (\log x)^2\), \(\dfrac{v'}{v} = \dfrac{2\log x}{x}\). Answer: \[(\log x)^x\left[\log(\log x) + \dfrac{1}{\log x}\right] + x^{\log x}\cdot\dfrac{2\log x}{x}\]
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Exercise 5.5, Question 8
\((\sin x)^x + \sin^{-1}\sqrt x\)
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\(u = (\sin x)^x\): \(\dfrac{u'}{u} = \log\sin x + x\cot x\). \[(\sin^{-1}\sqrt x)' = \dfrac{1}{\sqrt{1 - x}} \cdot \dfrac{1}{2\sqrt x}\] Answer: \[(\sin x)^x(x\cot x + \log\sin x) + \dfrac{1}{2\sqrt{x - x^2}}\]
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Exercise 5.5, Question 9
\(x^{\sin x} + (\sin x)^{\cos x}\)
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\(u = x^{\sin x}\): \[\dfrac{u'}{u} = \cos x\log x + \dfrac{\sin x}{x}\] \(v = (\sin x)^{\cos x}\): \[\dfrac{v'}{v} = -\sin x\log\sin x + \cos x\cot x\] Answer: \[x^{\sin x}\left(\cos x\log x + \dfrac{\sin x}{x}\right) + (\sin x)^{\cos x}(\cos x\cot x - \sin x\log\sin x)\]
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Exercise 5.5, Question 10
\(x^{x\cos x} + \dfrac{x^2 + 1}{x^2 - 1}\)
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\(u = x^{x\cos x}\): \(\log u = x\cos x\log x\), \[\dfrac{u'}{u} = \cos x\log x - x\sin x\log x + \cos x\] \[\begin{aligned}\left(\dfrac{x^2 + 1}{x^2 - 1}\right)' &= \dfrac{2x(x^2 - 1) - 2x(x^2 + 1)}{(x^2 - 1)^2} \\ &= -\dfrac{4x}{(x^2 - 1)^2}\end{aligned}\] Answer: \[x^{x\cos x}[\cos x(1 + \log x) - x\sin x\log x] - \dfrac{4x}{(x^2 - 1)^2}\]
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Exercise 5.5, Question 11
\((x\cos x)^x + (x\sin x)^{\frac1x}\)
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\(u = (x\cos x)^x\): \[\begin{aligned}\dfrac{u'}{u} &= \log(x\cos x) + x\left(\tfrac1x - \tan x\right) \\ &= \log(x\cos x) + 1 - x\tan x\end{aligned}\] \(v = (x\sin x)^{1/x}\): \[\begin{aligned}\dfrac{v'}{v} &= -\dfrac{\log(x\sin x)}{x^2} + \dfrac1x\left(\dfrac1x + \cot x\right) \\ &= \dfrac{1 + x\cot x - \log(x\sin x)}{x^2}\end{aligned}\] Answer: \[(x\cos x)^x[1 - x\tan x + \log(x\cos x)] + (x\sin x)^{\frac1x}\cdot\dfrac{x\cot x + 1 - \log(x\sin x)}{x^2}\]
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Exercise 5.5, Question 12
Find \(\dfrac{dy}{dx}\): \(x^y + y^x = 1\)
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\[(x^y)' = x^y\left(y' \log x + \dfrac{y}{x}\right)\], \[(y^x)' = y^x\left(\log y + \dfrac{x}{y}y'\right)\] Collect \(y'\): \[y'(x^y\log x + xy^{x-1}) = -(yx^{y-1} + y^x\log y)\] Answer: \[\dfrac{dy}{dx} = -\dfrac{yx^{y-1} + y^x\log y}{x^y\log x + xy^{x-1}}\]
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Exercise 5.5, Question 13
Find \(\dfrac{dy}{dx}\): \(y^x = x^y\)
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Take logs: \(x\log y = y\log x\). \[\log y + \dfrac{x}{y}y' = y'\log x + \dfrac{y}{x}\], so \[y'\left(\dfrac{x}{y} - \log x\right) = \dfrac{y}{x} - \log y\] Answer: \[\dfrac{dy}{dx} = \dfrac{y(y - x\log y)}{x(x - y\log x)}\]
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Exercise 5.5, Question 14
Find \(\dfrac{dy}{dx}\): \((\cos x)^y = (\cos y)^x\)
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Logs: \(y\log\cos x = x\log\cos y\). \[y'\log\cos x - y\tan x = \log\cos y - x\tan y\,y'\] Answer: \[\dfrac{dy}{dx} = \dfrac{y\tan x + \log\cos y}{x\tan y + \log\cos x}\]
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Exercise 5.5, Question 15
Find \(\dfrac{dy}{dx}\): \(xy = e^{x - y}\)
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Logs: \(\log x + \log y = x - y\). \[\begin{aligned}&\dfrac1x + \dfrac{y'}{y} = 1 - y' \\ \Rightarrow\ &y'\left(\dfrac1y + 1\right) = 1 - \dfrac1x\end{aligned}\] Answer: \[\dfrac{dy}{dx} = \dfrac{y(x - 1)}{x(y + 1)}\]
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Exercise 5.5, Question 16
\(f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\). Find \(f'(x)\) and \(f'(1)\).
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\(\log f = \sum\log(1 + x^{2^k})\): \[f'(x) = f(x)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]\] At 1: \(f(1) = 16\) and the bracket is \(\tfrac12 + 1 + 2 + 4 = \tfrac{15}{2}\). Answer: \[f'(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\left[\dfrac{1}{1 + x} + \dfrac{2x}{1 + x^2} + \dfrac{4x^3}{1 + x^4} + \dfrac{8x^7}{1 + x^8}\right]\]; \(f'(1) = 120\)
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Exercise 5.5, Question 17
Differentiate \((x^2 - 5x + 8)(x^3 + 7x + 9)\) (i) by the product rule, (ii) by expanding, (iii) by logarithmic differentiation. Do they agree?
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(i) \[(2x - 5)(x^3 + 7x + 9) + (x^2 - 5x + 8)(3x^2 + 7)\] (ii) The product is \(x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72\); differentiate term by term. (iii) \[\dfrac{y'}{y} = \dfrac{2x - 5}{x^2 - 5x + 8} + \dfrac{3x^2 + 7}{x^3 + 7x + 9}\]; multiplying by y gives the expression in (i). All three expand to the same polynomial. Answer: Yes: \(5x^4 - 20x^3 + 45x^2 - 52x + 11\) each time
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Exercise 5.5, Question 18
Show \(\dfrac{d}{dx}(uvw) = u'vw + uv'w + uvw'\) (i) by the product rule twice, (ii) by logarithmic differentiation.
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(i) \[\begin{aligned}(uv \cdot w)' &= (uv)'w + uvw' \\ &= (u'v + uv')w + uvw'\end{aligned}\] (ii) \(y = uvw\), \(\log y = \log u + \log v + \log w\) (where positive): \[\dfrac{y'}{y} = \dfrac{u'}{u} + \dfrac{v'}{v} + \dfrac{w'}{w}\]; multiply by \(y = uvw\). Answer: Shown both ways
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← Exercise 5.4 All Chapter 5 exercises Exercise 5.6 → Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in . Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log .