Question

If $y=\log _{\sin x}(\tan x)$, then $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$ is

• A. $\frac{4}{\log 2}$
• B. $\frac{-4}{\log 2}$
• C. $\frac{1}{\log 2}$
• D. None of these

Answer 1

Answer: (B)

We have,
$y=\log _{\sin x}(\tan x)=\frac{\log \tan x}{\log \sin x}$
$\therefore \frac{d y}{d x}=\frac{\log \sin x \cdot \frac{1}{\tan x} \cdot \sec ^2 x-\log \tan x \times \frac{1}{\sin x} \cos x}{(\log \sin x)^2}$
At $x=\frac{\pi}{4}$
$\left(\frac{d y}{d x}\right)_{x-\frac{\pi}{4}}=\frac{\left(\log \frac{1}{\sqrt{2}}\right)(\sqrt{2})^2}{\left[\log \left(\frac{1}{\sqrt{2}}\right)\right]^2}=\frac{(\sqrt{2})^2}{\log \left(\frac{1}{\sqrt{2}}\right)}=\frac{-2 \times 2}{\log 2}=\frac{-4}{\log 2}$