Question

Differentiate $\log \left(sinx\right)\left(logx\right)$

Answer 1

Let $y=\log \left(\sin x\right)\log x$

Differentiating above equation by Lebinitz rule, we have

$\frac{dy}{dx}=(\frac{1}{\sin x}.\cos x)\log x+\frac{1}{x}\log \left(\sin x\right)$

$\Rightarrow \frac{dy}{dx}=\frac{\cos x}{\sin x}\log x+\frac{\log \left(\sin x\right)}{x}$

$\Rightarrow \frac{dy}{dx}=\tan x\log x+\frac{\log \left(\sin x\right)}{x}$

$\Rightarrow \frac{dy}{dx}=\log \left(x^{\tan x}\right)+\log \left({\sin x}^{\frac{1}{x}}\right)$

$\Rightarrow \frac{dy}{dx}=\log \left(\left(x^{\tan x}\right)\left({\sin x}^{\frac{1}{x}}\right)\right)$

Hence $y^{\prime }=\log \left(\left(x^{\tan x}\right)\left({\sin x}^{\frac{1}{x}}\right)\right)$