Question

Differentiate ${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ with respect to ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right).$

Answer 1

Let $y=ta{n}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

$x=sin\theta$

$=ta{n}^{-1}\left(\frac{sin\theta }{\sqrt{1-si{n}^2\theta }}\right)=ta{n}^{-1}\left(\frac{sin\theta }{cos\theta }\right)=ta{n}^{-1}\left(tan\theta \right)=\theta$

$\frac{dy}{d\theta }=1$--- (1)

Let $z=si{n}^{-1}\left(2x\sqrt{1-x^2}\right)$

$=si{n}^{-1}\left(2sin\theta \sqrt{1-si{n}^{\theta }}\right)$

$=si{n}^{-1}\left(2sin\theta cos\theta \right)=si{n}^{-1}\left(sin2\theta \right)=2\theta$

$\frac{dz}{d\theta }=2$--- (2)

Now from (1) and (2),

$\frac{dy}{dz}=\frac{dy}{d\theta }\frac{d\theta }{dz}=1\times \frac{1}{2}=\frac{1}{2}.$