Question

If $y=\sin \left(\cos ^{-1}\left(\tan \left(\sec ^{-1} x\right)\right)\right), \quad-\sqrt{2} \leq x \leq \sqrt{2}$ then $\frac{d y}{d x}$ at $x=-1$ is

Answer 1

It is given that
$y=\sin \left(\cos ^{-1}\left(\tan \left(\sec ^{-1} x\right)\right)\right), \quad-\sqrt{2} \leq x \leq \sqrt{2}$
Let $sec^{- 1}x=\theta $
$\Rightarrow x=sec\theta $

$\Rightarrow y=sin\left(cos\right)^{- 1} \left(\right. tan \theta \left.\right)$
$=sin\left(cos\right)^{- 1} \left(\sqrt{x^{2} - 1}\right)$
Let $cos^{- 1}\sqrt{x^{2} - 1}=\phi$
$\Rightarrow cos\phi=\sqrt{x^{2} - 1}$
$\Rightarrow sin\phi=\sqrt{1 - cos^{2} \phi}$
$=\sqrt{1 - \left(x^{2} - 1\right)}=\sqrt{2 - x^{2}}$ $\therefore y=\sqrt{2 - x^{2}}$
$y^{'}=\frac{1}{2 \sqrt{2 - x^{2}}}\times \left(- 2 x\right)$ $=-\frac{x}{\sqrt{2 - x^{2}}}$
$\left(\frac{d y}{d x}\right)_{\left(x = - 1\right)}=\frac{- \left(- 1\right)}{\sqrt{2 - \left(1\right)^{2}}}=\frac{1}{1}=1$