Practicing Success
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 : |
$\sqrt{2}$ 1 $\frac{1}{\sqrt{2}}$ $\frac{1}{2\sqrt{2}}$ |
$\frac{1}{\sqrt{2}}$ |
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) $\frac{d(f(\cot x))}{d(g(cosec\,x)}=\frac{cosec\,xf'(\cot x)}{\cot xg'(cosec\,x)}$ $\left.\frac{d(f(\cot x))}{d(g(cosec\,x)}\right]_{x=π/4}=\frac{\sqrt{2}×f'(1)}{g'(\sqrt{2})}$ $=\sqrt{2}×2=2\sqrt{2}$ |