Question

Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1-x^2}-1}{x}\right)$ with respect to ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, when $x\ne 0.$

Answer 1

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

$y=ta{n}^{-1}\left(\frac{\sqrt{1+ta{n}^2\theta }-1}{tan\theta }\right)$

Put $x=tan\theta$

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

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

$\frac{dy}{d\theta }=\frac{1}{2}$

Let $z=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)=si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)=si{n}^{-1}\left(sin2\theta \right)=2\theta$

$\frac{dz}{d\theta }=2$

Now, $\frac{dy}{dz}=\frac{dy}{d\theta }.\frac{d\theta }{dz}=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}.$.