Question

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