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

The given equation is $y={\log }_{\sin x}\left(\tan x\right)$

 It can be written as $y=\frac{\log \tan x}{\log \sin x}$

Differentiating with respect to $x$,

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(\mathrm{logsin}x\right)\left(\frac{{\sec }^{2}x}{\tan x}\right)-\left(\mathrm{logtan}x\right)\left({\displaystyle \frac{\cos x}{\sin x}}\right)}{(\mathrm{logsin}x{)}^2} \\ \frac{dy}{dx}=\frac{\left(\mathrm{logsin}x\right)\left(\frac{{\sec }^{2}x}{\tan x}\right)-\left(\mathrm{logtan}x\right)\left({\displaystyle cotx}\right)}{(\mathrm{logsin}x{)}^2} \end{gathered}$$

At $x=\frac{\mathrm{\pi }}{4}$,

$$\begin{gathered} {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{logsin}\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)\left(\frac{{\sec }^2\left(\frac{\mathrm{\pi }}{4}\right)}{\tan \left(\frac{\mathrm{\pi }}{4}\right)}\right)-\left(\mathrm{logtan}\left(\frac{\mathrm{\pi }}{4}\right)\right){\displaystyle cot\left(\frac{\mathrm{\pi }}{4}\right)}}{{\left(\mathrm{logsin}\left(\frac{\mathrm{\pi }}{4}\right)\right)}^2} \\ {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=\frac{\log {\displaystyle \frac{1}{\sqrt{2}}}·2-0}{{\left(\log \left({\displaystyle \frac{1}{\sqrt{2}}}\right)\right)}^2} \\ {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=\frac{2}{\log \left(\frac{1}{\sqrt{2}}\right)} \\ {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=\frac{2}{{\left(\log \left(\sqrt{2}\right)\right)}^{-1}} \\ {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=-\frac{2}{\log \left(\sqrt{2}\right)} \\ {\left(\frac{dy}{dx}\right)}_{at x=\frac{\mathrm{\pi }}{4}}=-\frac{4}{\log 2} \end{gathered}$$

The value of $\frac{dy}{dx}$ at $x=\frac{\mathrm{\pi }}{4}$ is $-\frac{4}{\log 2}$.

Therefore, the correct answer is option $\left(3\right)$.