Let $f(x)=\cos \left[\cot ^{-1}\left( \frac{\cos x}{\sqrt{1-\cos 2x}}\right)\right]$ where $\frac{\pi}4<x<\frac{\pi}2$. Find $\frac{df(x)}{d\cot(x)}$

Question

Problem

Let $$f(x) = \cos \left[\cot ^{-1}\left( \frac{\cos x}{\sqrt {1 - \cos 2x}}\right)\right]$$ where $\frac{\pi}{4} < x < \frac{\pi}{2}$, then find the value of $$ \frac{d(f(x))}{d(\cot(x))}$$

Claimed Answer : $1$

My Try

Answer 1

You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cos\Bigg(cot^{-1}\Big(\frac{cot(x)}{\sqrt{2}}\Big)\Bigg)$$ (Note that I have corrected your calculation mistake and written $\sqrt{2}$ instead of $2$)

Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(\frac{y}{\sqrt{2}}))$. Apply the identity, $$cos(cot^{-1}(x))=\frac{x}{\sqrt{x^2+1}}$$ Now the function reduces to $\frac{y}{\sqrt{y^2+2}}$ and you can differentiate it easily.

Hope it helps:)