NCERT Exemplar Solutions Class 12 Maths Continuity and Differentiability — Short Answer Q25–43

NCERT Exemplar Problems, Class 12 Mathematics — Continuity and Differentiability, Short Answer questions 25 to 43. All 19 questions solved.

Nineteen differentiation exercises: exponential and logarithmic functions, nested trigonometric functions, logarithmic differentiation of powers and products, and then six inverse-trigonometric simplifications (Questions 37 to 43). The first group is the chain rule applied carefully. The last group is different: the interval printed beside each question is part of the problem, because it decides which branch of an inverse function applies and so whether the answer is $+$ or $-$.

Key insight. For $\tan^{-1}$ and $\cos^{-1}$ questions, do not differentiate first. Substitute, simplify to a single inverse function of a single angle, and check that the angle lies in the principal range ($(-\tfrac{\pi}{2},\tfrac{\pi}{2})$ for $\tan^{-1}$, $[0,\pi]$ for $\cos^{-1}$). Where it does not, adjust by a multiple of $\pi$ or reflect, and the derivative can change sign.

Question 25

Differentiate $2^{\cos^2 x}$ with respect to $x$.

Solution. Use $\dfrac{d}{dx}a^{u} = a^{u}\log a\cdot u’$ with $u = \cos^2 x$, so $u’ = -2\cos x\sin x = -\sin 2x$.

Question 26

Differentiate $\dfrac{8^x}{x^8}$ with respect to $x$.

Solution. By the quotient rule, $\dfrac{8^x\log 8\cdot x^8 – 8^x\cdot 8x^7}{x^{16}}$. Dividing numerator and denominator by $x^8$ gives the compact form below.

Question 27

Differentiate $\log\left(x+\sqrt{x^2+a}\right)$ with respect to $x$.

Solution. The derivative of the inside is $1 + \dfrac{x}{\sqrt{x^2+a}} = \dfrac{\sqrt{x^2+a}+x}{\sqrt{x^2+a}}$. Dividing by the inside $x+\sqrt{x^2+a}$, the common factor cancels.

Question 28

Differentiate $\log\left[\log\left(\log x^5\right)\right]$ with respect to $x$.

Solution. Peel the layers from outside in. The outer log gives $\dfrac{1}{\log(\log x^5)}$, the middle one gives $\dfrac{1}{\log x^5}$, and the innermost uses $\log x^5 = 5\log x$, whose derivative is $\dfrac{5}{x}$.

Question 29

Differentiate $\sin\sqrt{x}+\cos^2\sqrt{x}$ with respect to $x$.

Solution. With $\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}$: the first term gives $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$, and the second gives $2\cos\sqrt{x}\,(-\sin\sqrt{x})\cdot\dfrac{1}{2\sqrt{x}} = -\dfrac{\sin 2\sqrt{x}}{2\sqrt{x}}$.

Question 30

Differentiate $\sin^n(ax^2+bx+c)$ with respect to $x$.

Solution. This is a power of a sine of a quadratic: bring down $n$, lower the power, multiply by the derivative of the sine, then by the derivative of the quadratic.

Question 31

Differentiate $\cos\left(\tan\sqrt{x+1}\right)$ with respect to $x$.

Solution. Four layers: cosine, tangent, square root, and $x+1$. The derivatives multiply: $-\sin(\tan\sqrt{x+1})\cdot\sec^2\sqrt{x+1}\cdot\dfrac{1}{2\sqrt{x+1}}$.

Question 32

Differentiate $\sin x^2+\sin^2 x+\sin^2(x^2)$ with respect to $x$.

Solution. Three different terms. $\sin x^2$ gives $2x\cos x^2$. $\sin^2 x$ gives $2\sin x\cos x = \sin 2x$. $\sin^2(x^2)$ gives $2\sin x^2\cos x^2\cdot 2x = 2x\sin 2x^2$. The key prints the first term as “$2x\cos(x)^2$”, a typesetting slip; the term is $2x\cos(x^2)$.

Question 33

Differentiate $\sin^{-1}\left(\dfrac{1}{\sqrt{x+1}}\right)$ with respect to $x$.

Solution. The derivative of $\sin^{-1}u$ is $\dfrac{u’}{\sqrt{1-u^2}}$. Here $u = (x+1)^{-1/2}$, so $u’ = -\dfrac12(x+1)^{-3/2}$ and $1-u^2 = \dfrac{x}{x+1}$. Hence the result is $-\dfrac{1}{2(x+1)^{3/2}}\sqrt{\dfrac{x+1}{x}}$, which simplifies as below (the function is defined for $x>0$).

Question 34

Differentiate $(\sin x)^{\cos x}$ with respect to $x$.

Solution. The variable is in both base and exponent, so take logarithms: $\log y = \cos x\,\log\sin x$. Differentiating, $\dfrac{y’}{y} = -\sin x\log\sin x + \cos x\cdot\dfrac{\cos x}{\sin x}$. Multiply by $y$.

Question 35

Differentiate $\sin^m x\cdot\cos^n x$ with respect to $x$.

Solution. Use the product rule, or take logs: $\dfrac{y’}{y} = m\cot x – n\tan x$. The key prints the factor as “$\sin^{mx}$”, a typesetting slip for $\sin^m x$.

Question 36

Differentiate $(x+1)^2(x+2)^3(x+3)^4$ with respect to $x$.

Solution. Logarithmic differentiation is quicker than the product rule: $\dfrac{y’}{y} = \dfrac{2}{x+1}+\dfrac{3}{x+2}+\dfrac{4}{x+3}$. Putting the right side over a common denominator, the numerator is $2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2) = 9x^2+34x+29$. Multiplying by $y$ cancels one factor from each bracket.

Question 37

Differentiate $\cos^{-1}\left(\dfrac{\sin x+\cos x}{\sqrt{2}}\right)$, $-\dfrac{\pi}{4}<x<\dfrac{\pi}{4}$, with respect to $x$.

Solution. Since $\dfrac{\sin x+\cos x}{\sqrt2} = \cos\left(x-\dfrac{\pi}{4}\right)$, the function is $\cos^{-1}\cos u$ with $u = x-\dfrac{\pi}{4}\in\left(-\dfrac{\pi}{2},0\right)$. This $u$ lies outside $[0,\pi]$, so $\cos^{-1}\cos u = -u$, not $u$. Hence $y = \dfrac{\pi}{4}-x$.

Question 38

Differentiate $\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right)$, $-\dfrac{\pi}{4}<x<\dfrac{\pi}{4}$, with respect to $x$.

Solution. Using $1-\cos x = 2\sin^2\tfrac x2$ and $1+\cos x = 2\cos^2\tfrac x2$, the square root is $\left|\tan\tfrac x2\right|$, not $\tan\tfrac x2$. As $\left|\tfrac x2\right|<\tfrac{\pi}{8}$ lies in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, $y = \tan^{-1}\left|\tan\tfrac x2\right| = \dfrac{|x|}{2}$.

So $y’ = \dfrac12$ for $0<x<\dfrac{\pi}{4}$ and $y’ = -\dfrac12$ for $-\dfrac{\pi}{4}<x<0$; the function has a corner at $x=0$ and is not differentiable there. NCERT’s key prints only $\tfrac12$, which is correct on $\left(0,\tfrac{\pi}{4}\right)$ alone; the printed answer is incomplete for the stated interval.

Question 39

Differentiate $\tan^{-1}(\sec x+\tan x)$, $-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$, with respect to $x$.

Solution. $\sec x+\tan x = \dfrac{1+\sin x}{\cos x} = \tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$. For $x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ the angle $\tfrac{\pi}{4}+\tfrac x2$ lies in $\left(0,\tfrac{\pi}{2}\right)$, inside the principal range, so $y = \dfrac{\pi}{4}+\dfrac{x}{2}$.

Question 40

Differentiate $\tan^{-1}\left(\dfrac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$, $-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$ and $\dfrac{a}{b}\tan x>-1$, with respect to $x$.

Solution. Divide top and bottom by $b\cos x$: the fraction becomes $\dfrac{\frac ab-\tan x}{1+\frac ab\tan x} = \tan(\alpha-x)$ with $\alpha = \tan^{-1}\dfrac ab$. The condition $\dfrac ab\tan x>-1$ makes the denominator positive, which keeps $\alpha-x$ in the principal range, so $y = \tan^{-1}\dfrac ab – x$.

Question 41

Differentiate $\sec^{-1}\left(\dfrac{1}{4x^3-3x}\right)$, $0<x<\dfrac{1}{\sqrt{2}}$, with respect to $x$.

Solution. $\sec^{-1}\dfrac1u = \cos^{-1}u$, so $y = \cos^{-1}(4x^3-3x)$. Put $x = \cos\theta$ with $\theta\in\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)$; then $4x^3-3x = \cos3\theta$ and $3\theta\in\left(\tfrac{3\pi}{4},\tfrac{3\pi}{2}\right)$. Now $\cos^{-1}\cos3\theta = 3\theta$ only when $3\theta\le\pi$, that is $x\ge\tfrac12$.

For $\tfrac12<x<\tfrac{1}{\sqrt2}$: $y = 3\cos^{-1}x$, so $y’ = -\dfrac{3}{\sqrt{1-x^2}}$. For $0<x<\tfrac12$: $3\theta>\pi$, so $y = 2\pi-3\cos^{-1}x$ and $y’ = \dfrac{3}{\sqrt{1-x^2}}$.

NCERT’s key prints only $-\dfrac{3}{\sqrt{1-x^2}}$, right on $\left(\tfrac12,\tfrac{1}{\sqrt2}\right)$ but wrong on $\left(0,\tfrac12\right)$.

Question 42

Differentiate $\tan^{-1}\left(\dfrac{3a^2x-x^3}{a^3-3ax^2}\right)$, $-\dfrac{1}{\sqrt3}<\dfrac{x}{a}<\dfrac{1}{\sqrt3}$, with respect to $x$.

Solution. Put $x = a\tan\theta$. The fraction becomes $\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} = \tan3\theta$. The condition gives $\theta\in\left(-\tfrac{\pi}{6},\tfrac{\pi}{6}\right)$, so $3\theta\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$, which is the principal range. So $y = 3\tan^{-1}\dfrac xa$.

Question 43

Differentiate $\tan^{-1}\left(\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$, $-1<x<1$, $x\ne0$, with respect to $x$.

Solution. Put $x^2 = \cos2\theta$ with $2\theta\in\left(0,\tfrac{\pi}{2}\right)$. Then $\sqrt{1+x^2} = \sqrt2\cos\theta$ and $\sqrt{1-x^2} = \sqrt2\sin\theta$, and the fraction is $\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta} = \tan\left(\dfrac{\pi}{4}+\theta\right)$. The angle lies in $\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)$, so $y = \dfrac{\pi}{4}+\dfrac12\cos^{-1}x^2$, and $y’ = \dfrac12\cdot\dfrac{-2x}{\sqrt{1-x^4}}$.

Common mistakes

• Differentiating an inverse trigonometric function before simplifying it. Substituting first turns Questions 39 to 43 into one-line derivatives.
• Ignoring the principal range. In Question 37, $\cos^{-1}\cos u = -u$ because $u$ is negative; writing $u$ gives $+1$ instead of $-1$.
• Dropping the absolute value. $\sqrt{\tan^2\frac x2} = \left|\tan\frac x2\right|$, which is why Question 38 has two branches.
• Assuming $\cos^{-1}\cos3\theta = 3\theta$. It holds only for $3\theta\le\pi$; Question 41 needs both branches.
• Mishandling the chain rule in nested functions (Questions 28, 31): count the layers, then write one factor per layer.

Practise next

• Exemplar Short Answer Q1–24, the continuity and differentiability questions just before these.
• Exemplar Short Answer Q44–64, parametric and implicit differentiation, the next step.
• NCERT textbook Exercise 5.2 for chain-rule practice.
• NCERT textbook Exercise 5.3 for inverse trigonometric derivatives.