NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.3. All 15 questions solved.

The exercise has two halves.

Questions 1 to 8: implicit differentiation. Differentiate both sides with respect to $x$, remembering that every $y$ carries a factor of $\tfrac{dy}{dx}$ by the chain rule, then collect those terms and divide.

Questions 9 to 15: inverse trigonometric functions. Each looks like a hard chain rule and becomes a one-liner after a substitution.

$$x = \tan\theta \ \text{ for } \ \frac{2x}{1+x^2}, \ \frac{1-x^2}{1+x^2}, \ \frac{3x-x^3}{1-3x^2}; \qquad x = \sin\theta \ \text{ for } \ 2x\sqrt{1-x^2}$$

Key insight. In the second half, the substitution turns the argument into a single trigonometric ratio of a multiple angle, so the inverse function cancels and only a multiple of $\theta$ remains. For instance $\sin^{-1}\left(\tfrac{2x}{1+x^2}\right)$ becomes $\sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x$, whose derivative is $\tfrac{2}{1+x^2}$ — no quotient rule at all.

The stated domain restriction on each question is exactly what makes the cancellation valid. It is why questions 9 and 13 have derivatives of opposite sign despite identical-looking arguments: $\sin^{-1}$ and $\cos^{-1}$ of the same thing differ by a constant, and the constant differentiates away leaving a sign.

Find $\dfrac{dy}{dx}$ in the following:

Question 1

$2x + 3y = \sin x$

Solution. Differentiate throughout, treating $y$ as a function of $x$:

$$2 + 3\frac{dy}{dx} = \cos x$$

Question 2

$2x + 3y = \sin y$

Solution. Now the right side also carries a $\tfrac{dy}{dx}$:

$$2 + 3\frac{dy}{dx} = \cos y\,\frac{dy}{dx} \quad\Longrightarrow\quad 2 = \frac{dy}{dx}(\cos y – 3)$$

Question 3

$ax + by^2 = \cos y$

Solution.

$$a + 2by\frac{dy}{dx} = -\sin y\,\frac{dy}{dx} \quad\Longrightarrow\quad a = -\frac{dy}{dx}(2by + \sin y)$$

Question 4

$xy + y^2 = \tan x + y$

Solution. The first term needs the product rule:

$$y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = \sec^2x + \frac{dy}{dx}$$

Collecting the derivative terms on one side:

$$\frac{dy}{dx}(x + 2y – 1) = \sec^2x – y$$

Question 5

$x^2 + xy + y^2 = 100$

Solution.

$$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx}(x + 2y) = -(2x + y)$$

Question 6

$x^3 + x^2y + xy^2 + y^3 = 81$

Solution. Three of the four terms need the product rule:

$$3x^2 + \left(2xy + x^2\frac{dy}{dx}\right) + \left(y^2 + 2xy\frac{dy}{dx}\right) + 3y^2\frac{dy}{dx} = 0$$

Grouping:

$$\frac{dy}{dx}\left(x^2 + 2xy + 3y^2\right) = -\left(3x^2 + 2xy + y^2\right)$$

Question 7

$\sin^2 y + \cos xy = \kappa$

Solution. The first term has a chain rule with $y$ inside; the second has a product $xy$ inside a cosine.

$$2\sin y\cos y\,\frac{dy}{dx} – \sin(xy)\left(y + x\frac{dy}{dx}\right) = 0$$

Using $2\sin y\cos y = \sin 2y$ and collecting:

$$\frac{dy}{dx}\big(\sin 2y – x\sin xy\big) = y\sin xy$$

Question 8

$\sin^2 x + \cos^2 y = 1$

Solution.

$$2\sin x\cos x – 2\cos y\sin y\,\frac{dy}{dx} = 0$$

Both double-angle forms appear:

$$\sin 2x = \sin 2y\,\frac{dy}{dx}$$

Question 9

$y = \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$

Solution. Put $x = \tan\theta$, so $\theta = \tan^{-1}x$. Then

$$\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$

For the principal branch this gives $y = 2\theta = 2\tan^{-1}x$, and differentiating is now trivial.

Question 10

$y = \tan^{-1}\left(\dfrac{3x – x^3}{1 – 3x^2}\right)$, $-\dfrac{1}{\sqrt3} < x < \dfrac{1}{\sqrt3}$

Solution. With $x = \tan\theta$ the argument is the triple-angle formula:

$$\frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta} = \tan 3\theta$$

The restriction $|x| < \tfrac{1}{\sqrt3}$ gives $|\theta| < \tfrac{\pi}{6}$, so $3\theta$ stays inside $\tan^{-1}$’s branch and $y = 3\theta = 3\tan^{-1}x$.

Question 11

$y = \cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$, $0 < x < 1$

Solution. With $x = \tan\theta$, $\dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$, so $y = 2\theta = 2\tan^{-1}x$.

Question 12

$y = \sin^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)$, $0 < x < 1$

Solution. Same substitution, but now the argument $\cos 2\theta$ sits inside $\sin^{-1}$. Use $\sin^{-1}u = \tfrac{\pi}{2} – \cos^{-1}u$:

$$y = \frac{\pi}{2} – 2\tan^{-1}x$$

The constant differentiates away, leaving the negative of question 11’s answer.

Question 13

$y = \cos^{-1}\left(\dfrac{2x}{1+x^2}\right)$, $-1 < x < 1$

Solution. As in question 9 the argument is $\sin 2\theta$, and $\cos^{-1}u = \tfrac{\pi}{2} – \sin^{-1}u$ gives

$$y = \frac{\pi}{2} – 2\tan^{-1}x$$

Question 14

$y = \sin^{-1}\left(2x\sqrt{1-x^2}\right)$, $-\dfrac{1}{\sqrt2} < x < \dfrac{1}{\sqrt2}$

Solution. Here the root calls for $x = \sin\theta$, giving $\sqrt{1-x^2} = \cos\theta$ and

$$2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta$$

The restriction $|x| < \tfrac{1}{\sqrt2}$ gives $|\theta| < \tfrac{\pi}{4}$, so $2\theta$ stays in the branch and $y = 2\theta = 2\sin^{-1}x$.

Question 15

$y = \sec^{-1}\left(\dfrac{1}{2x^2 – 1}\right)$, $0 < x < \dfrac{1}{\sqrt2}$

Solution. Since $\sec^{-1}\tfrac{1}{u} = \cos^{-1}u$, this is

$$y = \cos^{-1}\left(2x^2 – 1\right)$$

Now put $x = \cos\theta$: then $2x^2 – 1 = \cos 2\theta$, so $y = 2\theta = 2\cos^{-1}x$.

Common mistakes

• Forgetting $\tfrac{dy}{dx}$ when differentiating a term in $y$. Every $y$ carries one. Question 2 differs from question 1 only in that respect, and the answers look nothing alike.
• Missing the product rule on a mixed term. $xy$, $x^2y$ and $xy^2$ in questions 4 and 6 each need it.
• Trying to differentiate an inverse trigonometric expression directly. Questions 9 to 15 are all one line after substitution and several lines of quotient rule without it.
• Ignoring the stated domain. It is what makes $\sin^{-1}(\sin 2\theta) = 2\theta$ legitimate; outside it a $\pi$ correction would be needed and the derivative would change sign.
• Missing that $\sec^{-1}\tfrac1u = \cos^{-1}u$. Without it question 15 is much harder than it needs to be.
• Not noticing the sign relationship. Questions 11 and 12 have the same argument and opposite answers, because $\sin^{-1} + \cos^{-1} = \tfrac{\pi}{2}$.

Practise next

• Exercise 5.4 — exponential and logarithmic functions, where the chain rule continues.
• Exercise 2.2 — the same inverse trigonometric substitutions, used there to simplify rather than to differentiate.