Question

The differentiation of ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ w.r.t. x is

• A. $\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(0,\infty \right)$
• B. $-\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(0,\infty \right)$
• C. $\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(-\infty ,0\right)$
• D. $-\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(-\infty ,0\right)$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(0,\infty \right)$
D $-\frac{2}{1+x^2},\mathrm{for}\mathrm{all}x\in \left(-\infty ,0\right)$

Let
$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right).\mathrm{Putting}x=\tan \theta ,\mathrm{we}\mathrm{get}\Rightarrow y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)={\cos }^{-1}\left(\cos 2\theta \right)\mathrm{Case}I:\mathrm{When}x\in \left(0,\infty \right)x=\tan \theta \Rightarrow 0<\tan \theta <\infty \Rightarrow 0<\theta <\frac{\pi }{2}\Rightarrow 0<2\theta <\pi \therefore y={\cos }^{-1}\left(\cos 2\theta \right)\Rightarrow y=2\theta =2{\tan }^{-1}x\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{1+x^2}\mathrm{Case}\mathrm{II}:\mathrm{When}x\in \left(-\infty ,0\right)x=\tan \theta \Rightarrow -\infty <x<0\Rightarrow -\frac{\pi }{2}<\theta <0\Rightarrow -\pi <2\theta <0\therefore y={\cos }^{-1}\left(\cos 2\theta \right)\Rightarrow y={\cos }^{-1}\left(\cos (-2\theta )\right]\left(\because -2\theta \in (0,\pi )\right)\Rightarrow y=-2\theta =-2{\tan }^{-1}x\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2}{1+x^2}$