Class 12 Maths NCERT Solutions Chapter 5 Ex 5.3 – Implicit and Inverse Trig Differentiation

Differentiate both sides with respect to x: 2+3\dfrac{dy}{dx}=\cos x.

Differentiate both sides, applying the chain rule to \sin y: 2+3\dfrac{dy}{dx}=\cos y\dfrac{dy}{dx}.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}(3-\cos y)=-2.

Differentiate both sides: a+2by\dfrac{dy}{dx}=-\sin y\dfrac{dy}{dx}.

Collect terms: \dfrac{dy}{dx}(2by+\sin y)=-a.

Differentiate xy with the product rule and y^2 with the chain rule: \left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=\sec^2x+\dfrac{dy}{dx}.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}(x+2y-1)=\sec^2x-y.

Differentiate each term: 2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}(x+2y)=-(2x+y).

Differentiate each term using the product rule on x^2y and xy^2: 3x^2+\left(2xy+x^2\dfrac{dy}{dx}\right)+\left(y^2+2xy\dfrac{dy}{dx}\right)+3y^2\dfrac{dy}{dx}=0.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}(x^2+2xy+3y^2)=-(3x^2+2xy+y^2).

Differentiate \sin^2y by the chain rule and \cos(xy) by the chain rule combined with the product rule on xy: 2\sin y\cos y\dfrac{dy}{dx}-\sin(xy)\left(y+x\dfrac{dy}{dx}\right)=0.

Collect the \dfrac{dy}{dx} terms: \dfrac{dy}{dx}\big(2\sin y\cos y-x\sin(xy)\big)=y\sin(xy).

Using the double-angle identity 2\sin y\cos y=\sin2y, this simplifies to \dfrac{dy}{dx}\big(\sin2y-x\sin(xy)\big)=y\sin(xy).

Differentiate both terms by the chain rule: 2\sin x\cos x+2\cos y(-\sin y)\dfrac{dy}{dx}=0.

Using the double-angle identities 2\sin x\cos x=\sin2x and 2\sin y\cos y=\sin2y: \sin2x-\sin2y\cdot\dfrac{dy}{dx}=0.

Substitute x=\tan\theta, so \theta=\tan^{-1}x. Then \dfrac{2x}{1+x^2}=\dfrac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta (double-angle identity).

So y=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x.

Differentiate: \dfrac{dy}{dx}=2\cdot\dfrac{1}{1+x^2}.

Substitute x=\tan\theta, so \theta=\tan^{-1}x. Then \dfrac{3x-x^3}{1-3x^2}=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta (triple-angle identity).

So y=\tan^{-1}(\tan3\theta)=3\theta=3\tan^{-1}x.

Differentiate: \dfrac{dy}{dx}=3\cdot\dfrac{1}{1+x^2}.

Substitute x=\tan\theta, so \theta=\tan^{-1}x. Then \dfrac{1-x^2}{1+x^2}=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta.

So y=\cos^{-1}(\cos2\theta)=2\theta=2\tan^{-1}x.

Differentiate: \dfrac{dy}{dx}=2\cdot\dfrac{1}{1+x^2}.

Substitute x=\tan\theta. Then \dfrac{1-x^2}{1+x^2}=\cos2\theta=\sin\left(\dfrac{\pi}{2}-2\theta\right).

So y=\sin^{-1}\left[\sin\left(\dfrac{\pi}{2}-2\theta\right)\right]=\dfrac{\pi}{2}-2\theta=\dfrac{\pi}{2}-2\tan^{-1}x.

Differentiate (the constant \pi/2 vanishes): \dfrac{dy}{dx}=-2\cdot\dfrac{1}{1+x^2}.

Substitute x=\tan\theta. Then \dfrac{2x}{1+x^2}=\sin2\theta=\cos\left(\dfrac{\pi}{2}-2\theta\right).

So y=\cos^{-1}\left[\cos\left(\dfrac{\pi}{2}-2\theta\right)\right]=\dfrac{\pi}{2}-2\theta=\dfrac{\pi}{2}-2\tan^{-1}x.

Differentiate: \dfrac{dy}{dx}=-2\cdot\dfrac{1}{1+x^2}.

Substitute x=\sin\theta, so \theta=\sin^{-1}x. Then 2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin2\theta.

So y=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x.

Differentiate: \dfrac{dy}{dx}=2\cdot\dfrac{1}{\sqrt{1-x^2}}.

Substitute x=\cos\theta, so \theta=\cos^{-1}x. Then 2x^2-1=2\cos^2\theta-1=\cos2\theta, so \dfrac{1}{2x^2-1}=\dfrac{1}{\cos2\theta}=\sec2\theta.

So y=\sec^{-1}(\sec2\theta)=2\theta=2\cos^{-1}x.

Differentiate: \dfrac{dy}{dx}=2\cdot\left(\dfrac{-1}{\sqrt{1-x^2}}\right).