If $y={\left(\sin x\right)}^{\cot x}$, then $\frac{dy}{dx}=$

$y={\left(\sin x\right)}^{\cot x}$

$\log y=\cot x.\log \left(\sin x\right)$

$\frac{1}{y}.\frac{dy}{dx}=\left(\cot x\right)\frac{1}{\sin x}.\left(\cos x\right)+\log \left(\sin x\right)(-\text{cosec}{x}^{2}x)$

$\frac{dy}{dx}=y[{\cot }^{2}x-{\text{cosec}}^{2}x.\log \left(\sin x\right)]$

$\frac{dy}{dx}={\left(\sin x\right)}^{\cot x}[{\cot }^{2}x-{\text{cosec}}^{2}x.\log \left(\sin x\right)]$