NCERT Exemplar Solutions Class 12 Maths Continuity and Differentiability — Short Answer Q44–64

NCERT Exemplar Problems, Class 12 Mathematics — Continuity and Differentiability, Short Answer questions 44 to 64. All 21 questions solved.

This stretch moves from explicit functions to curves given indirectly: parametric equations (Questions 44 to 51), a derivative with respect to another function (52, 53), implicit relations (54 to 57), and a run of proofs (58 to 64). Three tools do almost all the work: the chain rule in the form $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, logarithms to simplify before differentiating, and a good substitution to remove an awkward expression before any differentiation at all.

Key insight. Before differentiating a relation, ask whether it can be simplified first. A trigonometric substitution (Questions 47, 53), taking logarithms (49, 59, 60, 61), or solving for $x$ in terms of $y$ (59, 60, 62) usually turns a messy implicit derivative into a one-line calculation.

Question 44

Find $\dfrac{dy}{dx}$ for $x = t + \dfrac1t,\; y = t – \dfrac1t$.

Solution. Differentiate each with respect to $t$: $\dfrac{dx}{dt} = 1 – \dfrac1{t^2}$ and $\dfrac{dy}{dt} = 1 + \dfrac1{t^2}$. Dividing, $\dfrac{dy}{dx} = \dfrac{1 + 1/t^2}{1 – 1/t^2}$, and multiplying top and bottom by $t^2$ gives the form below.

Question 45

Find $\dfrac{dy}{dx}$ for $x = e^{\theta}\left(\theta + \dfrac1\theta\right),\; y = e^{-\theta}\left(\theta – \dfrac1\theta\right)$.

Solution. Use the product rule on each. For $x$: $\dfrac{dx}{d\theta} = e^{\theta}\left(\theta + \dfrac1\theta\right) + e^{\theta}\left(1 – \dfrac1{\theta^2}\right) = \dfrac{e^{\theta}}{\theta^2}\left(\theta^3 + \theta^2 + \theta – 1\right)$.

For $y$: $\dfrac{dy}{d\theta} = -e^{-\theta}\left(\theta – \dfrac1\theta\right) + e^{-\theta}\left(1 + \dfrac1{\theta^2}\right) = \dfrac{e^{-\theta}}{\theta^2}\left(-\theta^3 + \theta^2 + \theta + 1\right)$.

The factors $\theta^2$ cancel in the ratio and $e^{-\theta}/e^{\theta} = e^{-2\theta}$.

Question 46

Find $\dfrac{dy}{dx}$ for $x = 3\cos\theta – 2\cos^3\theta,\; y = 3\sin\theta – 2\sin^3\theta$.

Solution. $\dfrac{dx}{d\theta} = -3\sin\theta + 6\cos^2\theta\sin\theta = 3\sin\theta\,(2\cos^2\theta – 1) = 3\sin\theta\cos 2\theta$.

$\dfrac{dy}{d\theta} = 3\cos\theta – 6\sin^2\theta\cos\theta = 3\cos\theta\,(1 – 2\sin^2\theta) = 3\cos\theta\cos 2\theta$.

Both carry the factor $3\cos2\theta$, which cancels.

Question 47

Find $\dfrac{dy}{dx}$ for $\sin x = \dfrac{2t}{1+t^2},\; \tan y = \dfrac{2t}{1-t^2}$.

Solution. Put $t = \tan u$, so $u = \tan^{-1}t$. Then $\sin x = \dfrac{2\tan u}{1+\tan^2u} = \sin 2u$ and $\tan y = \dfrac{2\tan u}{1-\tan^2u} = \tan 2u$. For $|t|<1$ we have $u \in \left(-\dfrac\pi4, \dfrac\pi4\right)$, so $2u \in \left(-\dfrac\pi2, \dfrac\pi2\right)$, which lies in the principal range of both $\sin^{-1}$ and $\tan^{-1}$. Hence $x = 2u$ and $y = 2u$, that is $x = y = 2\tan^{-1}t$.

Both equal the same function of $t$, so their derivatives are equal and the ratio is $1$.

Question 48

Find $\dfrac{dy}{dx}$ for $x = \dfrac{1+\log t}{t^2},\; y = \dfrac{3+2\log t}{t}$.

Solution. By the quotient rule, $\dfrac{dx}{dt} = \dfrac{\frac1t\cdot t^2 – (1+\log t)\,2t}{t^4} = \dfrac{-(1+2\log t)}{t^3}$.

Likewise $\dfrac{dy}{dt} = \dfrac{\frac2t\cdot t – (3+2\log t)}{t^2} = \dfrac{-(1+2\log t)}{t^2}$.

The common factor $-(1+2\log t)$ cancels, leaving $t^3/t^2$.

Question 49

If $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$, prove that $\dfrac{dy}{dx} = \dfrac{-y\log x}{x\log y}$.

Solution. Differentiate directly: $\dfrac{dx}{dt} = -2\sin2t\; e^{\cos2t} = -2x\sin 2t$ and $\dfrac{dy}{dt} = 2\cos2t\; e^{\sin2t} = 2y\cos2t$. So $\dfrac{dy}{dx} = -\dfrac{y\cos2t}{x\sin2t}$.

Now $\log x = \cos2t$ and $\log y = \sin2t$ from the definitions, so substituting gives $-\dfrac{y\log x}{x\log y}$.

Question 50

If $x = a\sin 2t\,(1+\cos2t)$ and $y = b\cos2t\,(1-\cos2t)$, show that $\left(\dfrac{dy}{dx}\right)_{t=\pi/4} = \dfrac ba$.

Solution. By the product rule, $\dfrac{dx}{dt} = a\left[2\cos2t\,(1+\cos2t) – 2\sin^22t\right]$ and $\dfrac{dy}{dt} = b\left[-2\sin2t\,(1-\cos2t) + 2\sin2t\cos2t\right]$.

At $t = \pi/4$: $\cos2t = 0$ and $\sin2t = 1$. Then $\dfrac{dx}{dt} = a(0 – 2) = -2a$ and $\dfrac{dy}{dt} = b(-2 + 0) = -2b$. The ratio is $\dfrac{-2b}{-2a}$.

Question 51

If $x = 3\sin t – \sin3t,\; y = 3\cos t – \cos3t$, find $\dfrac{dy}{dx}$ at $t = \dfrac\pi3$.

Solution. $\dfrac{dx}{dt} = 3\cos t – 3\cos3t$ and $\dfrac{dy}{dt} = -3\sin t + 3\sin3t$. At $t = \pi/3$: $\cos t = \tfrac12$, $\cos\pi = -1$, $\sin t = \tfrac{\sqrt3}2$, $\sin\pi = 0$.

So $\dfrac{dx}{dt} = \tfrac32 + 3 = \tfrac92$ and $\dfrac{dy}{dt} = -\tfrac{3\sqrt3}2$. The ratio is $-\dfrac{3\sqrt3}{9} = -\dfrac{1}{\sqrt3}$.

Question 52

Differentiate $\dfrac{x}{\sin x}$ with respect to $\sin x$.

Solution. Let $u = \dfrac{x}{\sin x}$ and $v = \sin x$. The derivative of $u$ with respect to $v$ is $\dfrac{du/dx}{dv/dx}$. Here $\dfrac{du}{dx} = \dfrac{\sin x – x\cos x}{\sin^2x}$ and $\dfrac{dv}{dx} = \cos x$. Dividing, and splitting $\dfrac{\sin x – x\cos x}{\cos x} = \tan x – x$, gives the result.

Question 53

Differentiate $\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\tan^{-1}x$ when $x \ne 0$.

Solution. Put $x = \tan\theta$, so $\theta = \tan^{-1}x \in \left(-\dfrac\pi2, \dfrac\pi2\right)$. Then $\dfrac{\sqrt{1+x^2}-1}{x} = \dfrac{\sec\theta – 1}{\tan\theta} = \dfrac{1-\cos\theta}{\sin\theta} = \tan\dfrac\theta2$. Since $\dfrac\theta2 \in \left(-\dfrac\pi4, \dfrac\pi4\right)$, which lies in the principal range of $\tan^{-1}$, the function equals $\dfrac\theta2$.

So we differentiate $\dfrac\theta2$ with respect to $\theta$.

Question 54

Find $\dfrac{dy}{dx}$ if $\sin(xy) + \dfrac xy = x^2 – y$.

Solution. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

$$\cos(xy)\,(y + xy’) + \frac{y – xy’}{y^2} = 2x – y’.$$

Collect the terms in $y’$: $y’\left[x\cos(xy) – \dfrac{x}{y^2} + 1\right] = 2x – y\cos(xy) – \dfrac1y$. Multiplying top and bottom by $y^2$ clears the fractions.

Question 55

Find $\dfrac{dy}{dx}$ if $\sec(x+y) = xy$.

Solution. Differentiating, $\sec(x+y)\tan(x+y)\,(1 + y’) = y + xy’$. Bring the $y’$ terms together: $y’\left[\sec(x+y)\tan(x+y) – x\right] = y – \sec(x+y)\tan(x+y)$.

Question 56

Find $\dfrac{dy}{dx}$ if $\tan^{-1}(x^2+y^2) = a$.

Solution. Taking $\tan$ of both sides, $x^2 + y^2 = \tan a$, a constant. Differentiating, $2x + 2y\,y’ = 0$.

Question 57

Find $\dfrac{dy}{dx}$ if $(x^2+y^2)^2 = xy$.

Solution. Differentiate: $2(x^2+y^2)(2x + 2yy’) = y + xy’$, that is $4x(x^2+y^2) + 4y(x^2+y^2)\,y’ = y + xy’$. Collecting $y’$: $y’\left[4y(x^2+y^2) – x\right] = y – 4x(x^2+y^2)$. Expanding the brackets gives the form below.

Question 58

If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, show that $\dfrac{dy}{dx}\cdot\dfrac{dx}{dy} = 1$.

Solution. Differentiate with respect to $x$: $2ax + 2h(y + xy’) + 2byy’ + 2g + 2fy’ = 0$, so $\dfrac{dy}{dx} = -\dfrac{ax + hy + g}{hx + by + f}$.

Now differentiate the same relation with respect to $y$, treating $x$ as a function of $y$: $2axx’ + 2h(x’y + x) + 2by + 2gx’ + 2f = 0$, so $\dfrac{dx}{dy} = -\dfrac{hx + by + f}{ax + hy + g}$. The two expressions are reciprocals, so the product is $1$ (wherever both denominators are non-zero).

Question 59

If $x = e^{x/y}$, prove that $\dfrac{dy}{dx} = \dfrac{x-y}{x\log x}$.

Solution. Take logs: $\log x = \dfrac xy$, so $y = \dfrac{x}{\log x}$. Then $\dfrac{dy}{dx} = \dfrac{\log x – 1}{(\log x)^2}$.

Now substitute $\log x = \dfrac xy$ in the numerator and the denominator: $\dfrac{x/y – 1}{(\log x)\cdot (x/y)} = \dfrac{(x-y)/y}{x\log x/y} = \dfrac{x-y}{x\log x}$.

Question 60

If $y^x = e^{y-x}$, prove that $\dfrac{dy}{dx} = \dfrac{(1+\log y)^2}{\log y}$.

Solution. Taking logs, $x\log y = y – x$, so $x(1 + \log y) = y$ and $x = \dfrac{y}{1+\log y}$. Differentiating with respect to $y$:

$$\frac{dx}{dy} = \frac{(1+\log y) – y\cdot\frac1y}{(1+\log y)^2} = \frac{\log y}{(1+\log y)^2}.$$

Taking the reciprocal gives $\dfrac{dy}{dx}$.

Question 61

If $y = (\cos x)^{(\cos x)^{(\cos x)\cdots\infty}}$, show that $\dfrac{dy}{dx} = \dfrac{y^2\tan x}{y\log\cos x – 1}$.

Solution. The tower repeats itself, so the exponent is $y$ again: $y = (\cos x)^y$. Take logs: $\log y = y\log\cos x$. Differentiate:

$$\frac{y’}{y} = y’\log\cos x – y\tan x \;\Rightarrow\; y’\left(\frac1y – \log\cos x\right) = -y\tan x.$$

Multiplying through by $y$: $y'(1 – y\log\cos x) = -y^2\tan x$, which is the stated result after changing both signs.

Question 62

If $x\sin(a+y) + \sin a\cos(a+y) = 0$, prove that $\dfrac{dy}{dx} = \dfrac{\sin^2(a+y)}{\sin a}$.

Solution. Solve for $x$: $x = -\dfrac{\sin a\cos(a+y)}{\sin(a+y)} = -\sin a\cot(a+y)$. Differentiate with respect to $y$: $\dfrac{dx}{dy} = \sin a\csc^2(a+y) = \dfrac{\sin a}{\sin^2(a+y)}$. Then take the reciprocal.

Question 63

If $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x – y)$, prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1-y^2}{1-x^2}}$.

Solution. Put $x = \sin A$ and $y = \sin B$ with $A, B \in \left[-\dfrac\pi2, \dfrac\pi2\right]$, so $\sqrt{1-x^2} = \cos A$ and $\sqrt{1-y^2} = \cos B$ (both non-negative on this range). The relation becomes $\cos A + \cos B = a(\sin A – \sin B)$, that is

$$2\cos\frac{A+B}2\cos\frac{A-B}2 = 2a\cos\frac{A+B}2\sin\frac{A-B}2.$$

Cancelling $\cos\dfrac{A+B}2$ gives $\cot\dfrac{A-B}2 = a$, so $A – B = 2\cot^{-1}a$, a constant. That is, $\sin^{-1}x – \sin^{-1}y = $ constant. Differentiating, $\dfrac1{\sqrt{1-x^2}} – \dfrac1{\sqrt{1-y^2}}\dfrac{dy}{dx} = 0$.

Question 64

If $y = \tan^{-1}x$, find $\dfrac{d^2y}{dx^2}$ in terms of $y$ alone.

Solution. $\dfrac{dy}{dx} = \dfrac1{1+x^2}$. Since $x = \tan y$, this is $\dfrac1{1+\tan^2y} = \cos^2y$. Differentiate again with respect to $x$ using the chain rule: $\dfrac{d^2y}{dx^2} = -2\cos y\sin y\cdot\dfrac{dy}{dx} = -2\sin y\cos y\cdot\cos^2y$.

Common mistakes

• Forgetting the common factor before dividing (Questions 44 to 48). Compute $dx/dt$ and $dy/dt$ separately, factorise, and cancel; in Questions 46 and 48 the answer collapses to $\cot\theta$ and $t$ only after cancelling.
• Differentiating Question 47 as an implicit pair. Recognising $x = y = 2\tan^{-1}t$ gives the answer at once; the principal-value check ($|t|<1$) is what justifies it.
• Dropping the inner derivative in implicit differentiation. Every $y$-term needs a factor $y’$, and $\dfrac{x}{y}$ in Question 54 needs the quotient rule (Questions 54 to 57).
• Differentiating with respect to the wrong variable. In Questions 52 and 53 the answer is $\dfrac{du/dx}{dv/dx}$, not $du/dx$.
• Skipping the simplification step in the proofs. Solving for $x$ first (Questions 59, 60, 62) or substituting $x = \sin A$ (Question 63) is shorter and safer than differentiating the raw relation.

Practise next

• Exemplar Short Answer Q25–43, the inverse-trigonometric simplifications that come just before these.
• Exemplar Short Answer and Long Answer Q65–82, the next questions in this unit.
• NCERT textbook Exercise 5.3 for implicit and inverse-trigonometric differentiation.
• NCERT textbook Exercise 5.6 for parametric forms.