Complete solutions for Class 12 Maths (NCERT) Exercise 5.3 on Differentiation. Learn to find $\frac{dy}{dx}$ using Implicit Differentiation and simplify complex Inverse Trigonometric Functions with substitutions before differentiating.
This exercise requires finding $\frac{dy}{dx}$ using Implicit Differentiation (Questions 1-8) and applying trigonometric substitutions to simplify and differentiate Inverse Trigonometric Functions (Questions 9-15).
Implicit Differentiation (Questions 1 – 8)
When differentiating implicitly, remember that $\frac{d}{dx}(y) = \frac{dy}{dx}$ and use the Chain Rule for any function of $y$, such that $\frac{d}{dx}(f(y)) = f'(y) \cdot \frac{dy}{dx}$.
1. $2x + 3y = \sin x$
Differentiating both sides with respect to $x$:
$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$$
$$2 + 3 \frac{dy}{dx} = \cos x$$
$$3 \frac{dy}{dx} = \cos x – 2$$
$$\mathbf{\frac{dy}{dx} = \frac{\cos x – 2}{3}}$$
2. $2x + 3y = \sin y$
Differentiating both sides with respect to $x$:
$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$$
$$2 + 3 \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$$
Rearrange to solve for $\frac{dy}{dx}$:
$$2 = \cos y \frac{dy}{dx} – 3 \frac{dy}{dx}$$
$$2 = \frac{dy}{dx} (\cos y – 3)$$
$$\mathbf{\frac{dy}{dx} = \frac{2}{\cos y – 3}}$$
3. $ax + by^2 = \cos y$
Differentiating both sides with respect to $x$:
$$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$$
$$a + b (2y) \frac{dy}{dx} = (-\sin y) \frac{dy}{dx}$$
Rearrange to solve for $\frac{dy}{dx}$:
$$a = -\sin y \frac{dy}{dx} – 2by \frac{dy}{dx}$$
$$a = – \frac{dy}{dx} (\sin y + 2by)$$
$$\mathbf{\frac{dy}{dx} = – \frac{a}{\sin y + 2by}}$$
4. $xy + y^2 = \tan x + y$
Differentiating both sides with respect to $x$. Use the Product Rule for $xy$:
$$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)$$
$$\left(x \frac{dy}{dx} + y \cdot 1\right) + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$$
Group terms with $\frac{dy}{dx}$ on one side:
$$x \frac{dy}{dx} + 2y \frac{dy}{dx} – \frac{dy}{dx} = \sec^2 x – y$$
$$\frac{dy}{dx} (x + 2y – 1) = \sec^2 x – y$$
$$\mathbf{\frac{dy}{dx} = \frac{\sec^2 x – y}{x + 2y – 1}}$$
5. $x^2 + xy + y^2 = 100$
Differentiating both sides with respect to $x$. Use the Product Rule for $xy$:
$$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(100)$$
$$2x + \left(x \frac{dy}{dx} + y \cdot 1\right) + 2y \frac{dy}{dx} = 0$$
Group terms with $\frac{dy}{dx}$:
$$x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x – y$$
$$\frac{dy}{dx} (x + 2y) = -(2x + y)$$
$$\mathbf{\frac{dy}{dx} = – \frac{2x + y}{x + 2y}}$$
6. $x^3 + x^2y + xy^2 + y^3 = 81$
Differentiating term by term with respect to $x$. Use the Product Rule for $x^2y$ and $xy^2$:
$$\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81)$$
$$3x^2 + \left(x^2 \frac{dy}{dx} + y \cdot 2x\right) + \left(x \cdot 2y \frac{dy}{dx} + y^2 \cdot 1\right) + 3y^2 \frac{dy}{dx} = 0$$
Group terms with $\frac{dy}{dx}$:
$$\frac{dy}{dx} (x^2 + 2xy + 3y^2) = – (3x^2 + 2xy + y^2)$$
$$\mathbf{\frac{dy}{dx} = – \frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}}$$
7. $\sin^2 y + \cos(xy) = \kappa$ (where $\kappa$ is a constant)
Differentiating both sides with respect to $x$:
$$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$$
Use the Chain Rule for $\sin^2 y = (\sin y)^2$:
$$2 \sin y \cdot \cos y \cdot \frac{dy}{dx} + (-\sin(xy)) \cdot \frac{d}{dx}(xy) = 0$$
Using the identity $2 \sin y \cos y = \sin(2y)$ and the Product Rule for $xy$:
$$\sin(2y) \frac{dy}{dx} – \sin(xy) \left(x \frac{dy}{dx} + y \cdot 1\right) = 0$$
$$\sin(2y) \frac{dy}{dx} – x \sin(xy) \frac{dy}{dx} – y \sin(xy) = 0$$
Group terms with $\frac{dy}{dx}$:
$$\frac{dy}{dx} (\sin(2y) – x \sin(xy)) = y \sin(xy)$$
$$\mathbf{\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) – x \sin(xy)}}$$
8. $\sin^2 x + \cos^2 y = 1$
Differentiating both sides with respect to $x$:
$$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1)$$
$$2 \sin x \cdot \cos x + 2 \cos y \cdot (-\sin y) \cdot \frac{dy}{dx} = 0$$
Using the identities $2 \sin x \cos x = \sin(2x)$ and $2 \cos y \sin y = \sin(2y)$:
$$\sin(2x) – \sin(2y) \frac{dy}{dx} = 0$$
$$\sin(2x) = \sin(2y) \frac{dy}{dx}$$
$$\mathbf{\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}}$$
Inverse Trigonometric Functions (Questions 9 – 15)
These problems are best solved by using an appropriate trigonometric substitution to simplify the inner expression and then applying the derivative formula $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$, etc.
9. $y = \sin^{-1}\left(\frac{2x}{1 + x^2}\right), -1 < x < 1$
Substitution: Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
$$y = \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right)$$
Using the identity $\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$:
$$y = \sin^{-1}(\sin(2\theta))$$
Since $-1 < x < 1$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, so $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. Within this range, $\sin^{-1}(\sin(2\theta)) = 2\theta$.
$$y = 2\theta$$
Substituting back $\theta = \tan^{-1} x$:
$$y = 2 \tan^{-1} x$$
Differentiating with respect to $x$:
$$\mathbf{\frac{dy}{dx} = \frac{2}{1 + x^2}}$$
10. $y = \tan^{-1}\left(\frac{3x – x^3}{1 – 3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$
Substitution: Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
$$y = \tan^{-1}\left(\frac{3 \tan \theta – \tan^3 \theta}{1 – 3 \tan^2 \theta}\right)$$
Using the identity $\tan(3\theta) = \frac{3 \tan \theta – \tan^3 \theta}{1 – 3 \tan^2 \theta}$:
$$y = \tan^{-1}(\tan(3\theta))$$
Since $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$, we have $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, so $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$. Within this range, $\tan^{-1}(\tan(3\theta)) = 3\theta$.
$$y = 3\theta$$
Substituting back $\theta = \tan^{-1} x$:
$$y = 3 \tan^{-1} x$$
Differentiating with respect to $x$:
$$\mathbf{\frac{dy}{dx} = \frac{3}{1 + x^2}}$$
11. $y = \cos^{-1}\left(\frac{1 – x^2}{1 + x^2}\right), 0 < x < 1$
Substitution: Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
$$y = \cos^{-1}\left(\frac{1 – \tan^2 \theta}{1 + \tan^2 \theta}\right)$$
Using the identity $\cos(2\theta) = \frac{1 – \tan^2 \theta}{1 + \tan^2 \theta}$:
$$y = \cos^{-1}(\cos(2\theta))$$
Since $0 < x < 1$, we have $0 < \theta < \frac{\pi}{4}$, so $0 < 2\theta < \frac{\pi}{2}$. Within this range, $\cos^{-1}(\cos(2\theta)) = 2\theta$.
$$y = 2\theta$$
Substituting back $\theta = \tan^{-1} x$:
$$y = 2 \tan^{-1} x$$
Differentiating with respect to $x$:
$$\mathbf{\frac{dy}{dx} = \frac{2}{1 + x^2}}$$
12. $y = \sin^{-1}\left(\frac{1 – x^2}{1 + x^2}\right), 0 < x < 1$
Substitution: Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
$$y = \sin^{-1}\left(\frac{1 – \tan^2 \theta}{1 + \tan^2 \theta}\right)$$
Using the identity $\cos(2\theta) = \frac{1 – \tan^2 \theta}{1 + \tan^2 \theta}$:
$$y = \sin^{-1}(\cos(2\theta))$$
Using the complementary angle identity $\cos A = \sin\left(\frac{\pi}{2} – A\right)$:
$$y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} – 2\theta\right)\right)$$
Since $0 < 2\theta < \frac{\pi}{2}$, we have $0 < \frac{\pi}{2} – 2\theta < \frac{\pi}{2}$. Within this range, $\sin^{-1}(\sin A) = A$.
$$y = \frac{\pi}{2} – 2\theta$$
Substituting back $\theta = \tan^{-1} x$:
$$y = \frac{\pi}{2} – 2 \tan^{-1} x$$
Differentiating with respect to $x$:
$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) – 2 \frac{d}{dx}(\tan^{-1} x)$$
$$\mathbf{\frac{dy}{dx} = 0 – \frac{2}{1 + x^2} = – \frac{2}{1 + x^2}}$$
13. $y = \cos^{-1}\left(\frac{2x}{1 + x^2}\right), -1 < x < 1$
Substitution: Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
$$y = \cos^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right)$$
Using the identity $\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$:
$$y = \cos^{-1}(\sin(2\theta))$$
Using the complementary angle identity $\sin A = \cos\left(\frac{\pi}{2} – A\right)$:
$$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} – 2\theta\right)\right)$$
Since $-1 < x < 1$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, so $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. This means $0 < \frac{\pi}{2} – 2\theta < \pi$. Within this range, $\cos^{-1}(\cos A) = A$.
$$y = \frac{\pi}{2} – 2\theta$$
Substituting back $\theta = \tan^{-1} x$:
$$y = \frac{\pi}{2} – 2 \tan^{-1} x$$
Differentiating with respect to $x$:
$$\mathbf{\frac{dy}{dx} = – \frac{2}{1 + x^2}}$$
14. $y = \sin^{-1}(2x\sqrt{1 – x^2}), -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
Substitution: Let $x = \sin \theta$. Then $\theta = \sin^{-1} x$.
$$y = \sin^{-1}(2 \sin \theta \sqrt{1 – \sin^2 \theta})$$
Using the identity $\cos \theta = \sqrt{1 – \sin^2 \theta}$:
$$y = \sin^{-1}(2 \sin \theta \cos \theta)$$
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:
$$y = \sin^{-1}(\sin(2\theta))$$
Since $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, so $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. Within this range, $\sin^{-1}(\sin(2\theta)) = 2\theta$.
$$y = 2\theta$$
Substituting back $\theta = \sin^{-1} x$:
$$y = 2 \sin^{-1} x$$
Differentiating with respect to $x$:
$$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\sin^{-1} x)$$
$$\mathbf{\frac{dy}{dx} = \frac{2}{\sqrt{1 – x^2}}}$$
15. $y = \sec^{-1}\left(\frac{1}{2x^2 – 1}\right), 0 < x < \frac{1}{\sqrt{2}}$
Substitution: Let $x = \cos \theta$. Then $\theta = \cos^{-1} x$.
$$y = \sec^{-1}\left(\frac{1}{2 \cos^2 \theta – 1}\right)$$
Using the identity $\cos(2\theta) = 2 \cos^2 \theta – 1$:
$$y = \sec^{-1}\left(\frac{1}{\cos(2\theta)}\right)$$
Using the identity $\sec(A) = \frac{1}{\cos(A)}$:
$$y = \sec^{-1}(\sec(2\theta))$$
Since $0 < x < \frac{1}{\sqrt{2}}$, we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, so $\frac{\pi}{2} < 2\theta < \pi$.
The range of $\sec^{-1} A$ is $[0, \pi] – \{\frac{\pi}{2}\}$.
Since $\frac{\pi}{2} < 2\theta < \pi$, $\sec^{-1}(\sec(2\theta)) = 2\theta$.
$$y = 2\theta$$
Substituting back $\theta = \cos^{-1} x$:
$$y = 2 \cos^{-1} x$$
Differentiating with respect to $x$:
$$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\cos^{-1} x)$$
$$\mathbf{\frac{dy}{dx} = \frac{-2}{\sqrt{1 – x^2}}}$$