Question

If $y={\log }_{\sin x}\left(\tan x\right),\text{then}{\left(\frac{dy}{dx}\right)}_{\pi /4}$ is equal to

• A. $\frac{4}{\log 2}$
• B. $-4\log 2$
• C. $\frac{-4}{\log 2}$
• D. $4\log 2$

Answer 1

The correct option is C $\frac{-4}{\log 2}$

Solution :-

$y=lo{g}_{sinx}\left(tanx\right),$

$y=\frac{log\left(tanx\right)}{log\left(sinx\right)}$

$\frac{dy}{dx}=\frac{log\left(sinx\right).\frac{se{c}^{2}x}{tanx}-log\tan x.\frac{cosx}{sinx}}{\left[log\right(sinx){]}^2}$

Now $(\frac{dy}{dx}{)}_{\pi /4}=\frac{log\left(sin\pi /4\right)\frac{se{c}^2\pi /4}{tan\pi /4}-log\tan \pi /4.\frac{cos\pi /4}{sin\pi /4}}{\left[log\right(sin\pi /4){]}^2}$

$=\frac{log\left(\frac{1}{\sqrt{2}}\right).2-0}{\left(log\right(1/\sqrt{2}){)}^2}\Rightarrow \frac{2}{log(\sqrt{2}{)}^{-1}}$

$=\frac{-2}{log\sqrt{2}}$

$=\frac{-4}{log2}$