Question

If $y={\sin }^2\left({\tan }^{-1}\sqrt{\frac{1-x^2}{1+x^2}}\right),$ then $\frac{dy}{dx}$ =

• A. x
• B. -x
• C. 1
• D. -1

Answer 1

Answer: (B)

$y={\sin }^2\left({\tan }^{-1}\sqrt{\frac{1-x^2}{1+x^2}}\right)$
Put $x^2=\cos \theta \Rightarrow 2x\frac{dx}{d\theta }=-\sin \theta$
$\Rightarrow y={\sin }^2\left({\tan }^{-1}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\right)$
$={\sin }^2\left({\tan }^{-1}\sqrt{\frac{2{\sin }^2\left(\theta 2\right)}{2{\cos }^2\left(\theta 2\right)}}\right)$
$={\sin }^2\left({\tan }^{-1}\left(\tan \frac{\theta }{2}\right)\right)$
$\Rightarrow y={\sin }^2\left(\frac{\theta }{2}\right)$
Now, $\frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\left(2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right).\frac{1}{2}\times \left(\frac{-2x}{\sin \theta }\right)$
$=2\sin \frac{\theta }{2}\cos \frac{\theta }{2}.\frac{1}{2}\times \frac{-2x}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$
$=-x$