If $y=\log an \left(\frac{x}{2} ight)+\sin ^1(\cos x)$, then $\frac{d y}{d x}=$

Question

If $y=\log 	an \left(\frac{x}{2}ight)+\sin ^1(\cos x)$, then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

$\begin{aligned} & y=\log 	an \left(\frac{x}{2}ight)+\sin ^{-1}(\cos x) \ & =\log 	an \left(\frac{x}{2}ight)+\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-xight)ight]=\log 	an \left(\frac{x}{2}ight)+\frac{\pi}{2}-x \ & 	herefore \frac{d y}{d x}=\frac{1}{	an \left(\frac{x}{2}ight)} 	imes \sec ^2 	imes\left(\frac{x}{2}ight) 	imes\left(\frac{1}{2}ight)+0-1 \ & =\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} 	imes \frac{1}{\cos ^2 \frac{x}{2}} 	imes \frac{1}{2}-1 \ & =\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}-1=\frac{1}{\sin x}-1=\operatorname{cosec} x-1\end{aligned}$

If $y=\log \tan \left(\frac{x}{2}\right)+\sin ^1(\cos x)$, then $\frac{d y}{d x}=$

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