Question

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

Answer 1

Answer: 1

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

$\Rightarrow y=sin{\left(cos\right)}^{-1}(tan\theta )$
$=sin{\left(cos\right)}^{-1}\left(\sqrt{x^2-1}\right)$
Let $co{s}^{-1}\sqrt{x^2-1}=\varphi$
$\Rightarrow cos\varphi =\sqrt{x^2-1}$
$\Rightarrow sin\varphi =\sqrt{1-co{s}^2\varphi }$
$=\sqrt{1-\left(x^2-1\right)}=\sqrt{2-x^2}$ $\therefore y=\sqrt{2-x^2}$
$y^{{}^{\prime }}=\frac{1}{2\sqrt{2-x^2}}\times \left(-2x\right)$ $=-\frac{x}{\sqrt{2-x^2}}$
${\left(\frac{dy}{dx}\right)}_{\left(x=-1\right)}=\frac{-\left(-1\right)}{\sqrt{2-{\left(1\right)}^2}}=\frac{1}{1}=1$