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} } $
$= -x$