Question

Find $\frac{dy}{dx}$ for $y=\log \left(\tan \frac{x}{2}\right)+{\sin }^{-1}\left(\cos x\right)$.

Answer 1

Given,

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

or, $y=\log \left(\tan \frac{x}{2}\right)+{\sin }^{-1}\sin (\frac{\pi }{2}-x)$

or, $y=\log \left(\tan \frac{x}{2}\right)+\frac{\pi }{2}-x$

Now, differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{{\sec }^2\frac{x}{2}}{\tan \frac{x}{2}}.\frac{1}{2}-x$

or, $\frac{dy}{dx}=\frac{1}{2.\sin \frac{x}{2}.\cos \frac{x}{2}}-1$

or, $\frac{dy}{dx}=\frac{1}{\sin x}-1$

or, $\frac{dy}{dx}=cosecx-1$.