NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Substitution. Put \(\displaystyle x=\tan\theta\), so \(\displaystyle \theta=\tan^{-1}x\). With \(\displaystyle 0<x<1\), \[0<\theta<\frac{\pi}{4}. \]Using \(\displaystyle \dfrac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta\), \[y=\cos^{-1}(\cos 2\theta). \]Branch check: \(\displaystyle \cos^{-1}(\cos\alpha)=\alpha\) only for \(\displaystyle \alpha\in[0,\pi]\). Here \(\displaystyle 2\theta\in\left(0,\frac{\pi}{2}\right)\subset[0,\pi]\), so the cancellation is valid and \[y=2\theta=2\tan^{-1}x. \]Differentiating, \[\boxed{\ \frac{dy}{dx}=\frac{2}{1+x^{2}}\ } \]

Substitution. Put \(\displaystyle x=\tan\theta\), \(\displaystyle \theta=\tan^{-1}x\); with \(\displaystyle 0<x<1\) we get \(\displaystyle 0<\theta<\frac{\pi}{4}\), so \(\displaystyle 0<2\theta<\frac{\pi}{2}\).As in the previous question the argument is \(\displaystyle \cos 2\theta\), but here it sits inside \(\displaystyle \sin^{-1}\), so convert the cosine to a sine using \(\displaystyle \cos 2\theta=\sin\left(\frac{\pi}{2}-2\theta\right)\): \[y=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right). \]Branch check: \(\displaystyle \sin^{-1}(\sin\alpha)=\alpha\) requires \(\displaystyle \alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Since \(\displaystyle 0<2\theta<\frac{\pi}{2}\), we have \(\displaystyle \frac{\pi}{2}-2\theta\in\left(0,\frac{\pi}{2}\right)\), so the cancellation is valid: \[y=\frac{\pi}{2}-2\theta=\frac{\pi}{2}-2\tan^{-1}x. \]Differentiating (the constant \(\displaystyle \frac{\pi}{2}\) contributes nothing, and the minus sign survives): \[\boxed{\ \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ } \]

Substitution. Put \(\displaystyle x=\tan\theta\), \(\displaystyle \theta=\tan^{-1}x\); with \(\displaystyle -1<x<1\), \[-\frac{\pi}{4}<\theta<\frac{\pi}{4},\qquad\text{so}\qquad -\frac{\pi}{2}<2\theta<\frac{\pi}{2}. \]Then \(\displaystyle \dfrac{2x}{1+x^{2}}=\sin 2\theta\), which must be turned into a cosine because the outer function is \(\displaystyle \cos^{-1}\): \[\sin 2\theta=\cos\left(\frac{\pi}{2}-2\theta\right),\qquad y=\cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right). \]Branch check: \(\displaystyle \cos^{-1}(\cos\alpha)=\alpha\) needs \(\displaystyle \alpha\in[0,\pi]\). From \(\displaystyle -\frac{\pi}{2}<2\theta<\frac{\pi}{2}\) we get \(\displaystyle \frac{\pi}{2}-2\theta\in(0,\pi)\), so the cancellation holds and \[y=\frac{\pi}{2}-2\tan^{-1}x. \]Differentiating, \[\boxed{\ \frac{dy}{dx}=-\frac{2}{1+x^{2}}\ } \]

Substitution. The factor \(\displaystyle \sqrt{1-x^{2}}\) suggests a sine substitution: put \(\displaystyle x=\sin\theta\), so \(\displaystyle \theta=\sin^{-1}x\). The restriction \(\displaystyle -\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\) gives \[-\frac{\pi}{4}<\theta<\frac{\pi}{4}. \]On this interval \(\displaystyle \cos\theta>0\), so \(\displaystyle \sqrt{1-x^{2}}=\sqrt{1-\sin^{2}\theta}=|\cos\theta|=\cos\theta\) \(\displaystyle -\) the positive root is the one to take, and this is where the restriction earns its keep. Hence \[2x\sqrt{1-x^{2}}=2\sin\theta\cos\theta=\sin 2\theta,\qquad y=\sin^{-1}(\sin 2\theta). \]Branch check: \(\displaystyle 2\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), which lies in the principal range of \(\displaystyle \sin^{-1}\), so \[y=2\theta=2\sin^{-1}x. \]Differentiating with \(\displaystyle \frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}\): \[\boxed{\ \frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}\ } \]

Substitution. Since \(\displaystyle \mathrm{sec}^{-1}\) undoes a secant, first make the argument a secant. Put \(\displaystyle x=\cos\theta\), so \(\displaystyle \theta=\cos^{-1}x\). The restriction \(\displaystyle 0<x<\frac{1}{\sqrt{2}}\) gives \[\frac{\pi}{4}<\theta<\frac{\pi}{2}. \]Using \(\displaystyle 2\cos^{2}\theta-1=\cos 2\theta\), \[\frac{1}{2x^{2}-1}=\frac{1}{\cos 2\theta}=\sec 2\theta,\qquad y=\mathrm{sec}^{-1}(\sec 2\theta). \]Branch check: \(\displaystyle \mathrm{sec}^{-1}(\sec\alpha)=\alpha\) requires \(\displaystyle \alpha\in[0,\pi]\) with \(\displaystyle \alpha\ne\frac{\pi}{2}\). From \(\displaystyle \frac{\pi}{4}<\theta<\frac{\pi}{2}\) we get \(\displaystyle 2\theta\in\left(\frac{\pi}{2},\pi\right)\), which is inside the principal range (and note \(\displaystyle \cos 2\theta<0\) there, consistent with \(\displaystyle 2x^{2}-1<0\) for \(\displaystyle x<\frac{1}{\sqrt{2}}\)). Hence \[y=2\theta=2\cos^{-1}x. \]Differentiating with \(\displaystyle \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^{2}}}\): \[\boxed{\ \frac{dy}{dx}=-\frac{2}{\sqrt{1-x^{2}}}\ } \]