The derivative of f(cot x) with respect to g (cosec x) at $x=\frac{\pi}{4}$ (where $f'(1)=2g'(\sqrt{2})=4)$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $2\sqrt{2}$
$\frac{d}{dx}(f(\cot x))=f'(\cot x)(-cosec^2x)$ ...(1)
$\frac{d}{dx}(g(cosec\,x))=g'(cosec\,x)(-cosec\,x\cot x)$ ...(2)
$\frac{d(f(\cot x))}{d(g(cosec\,x)}=\frac{cosec\,xf'(\cot x)}{\cot xg'(cosec\,x)}$
$\left.\frac{d(f(\cot x))}{d(g(cosec\,x)}\right]_{x=π/4}=\frac{\sqrt{2}×f'(1)}{g'(\sqrt{2})}$
$=\sqrt{2}×2=2\sqrt{2}$