Discussion Forum

Derivative of  $\sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$ with respect to  $\cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ is 

Answer:

Explanation:

Let y = $\sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$

put  $t= \tan\theta\Rightarrow\theta=\tan^{-1}t$

 = $\sin^{-1}\left(\frac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}\right)=\sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right)$

 =$\sin^{-1}(\sin\theta)=\theta=\tan^{-1}t$

 and   $z=\cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)=\cos^{-1}\left(\frac{1}{\sqrt{1+\tan^{2}\theta}}\right)=\cos^{-1}(\cos \theta)$

$\theta=\tan^{-1}t$

 $\therefore$    $\frac{dy}{dz}=\frac{\frac{dy}{dt}}{\frac{dz}{st}}$

 $=\frac{\frac{1}{(1+t^{2})}}{\frac{1}{(1+t^{2})}}=1$