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#### If $\left ( \cot x \right )^{y}= \left ( \cot y \right )^{x}$ find $\frac{dy}{dx}$.Option: 1 $\frac{\log (\cot y)+x\left(\frac{1}{\sin x \cos x}\right)}{\log (\cot x)+x\left(\frac{1}{\sin y \cos y}\right)}$Option: 2 $\frac{\log (\cot y)-y\left(\frac{1}{\sin x \cos x}\right)}{\log (\cot x)+x\left(\frac{1}{\sin y \cos y}\right)}$Option: 3 $\frac{\log (\cot y)-y\left(\frac{1}{\sin x \cos x}\right)}{\log (\cot x)-x\left(\frac{1}{\sin y \cos y}\right)}$Option: 4 $\frac{\log (\cot y)+y\left(\frac{1}{\sin x \cos x}\right)}{\log (\cot x)+x\left(\frac{1}{\sin y \cos y}\right)}$

The given term is:

$(\cot x)^y=(\cot y)^x$

Taking logs on both sides,

$y \log (\cot x)=x \log (\cot y)$

Then differentiating both sides w.r.t. x,
$y \cdot \frac{1}{\cot x} \cdot \frac{d}{d x}(\cot x)+\log (\cot x) \cdot \frac{d y}{d x}=x \cdot \frac{1}{\cot y} \cdot \frac{d}{d x}(\cot y) \frac{d y}{d x}+\log (\cot y) \cdot 1$
$y \cdot \frac{1}{\cot x} \cdot\left(-\csc ^2 x\right)+\log (\cot x) \cdot \frac{d y}{d x}=x \cdot \frac{1}{\cot y} \cdot\left(-\csc ^2 y\right) \frac{d y}{d x}+\log (\cot y)$
$-y \cdot \frac{\csc ^2 x}{\cot x}+\log (\cot x) \frac{d y}{d x}=x \cdot\left(\frac{-\csc ^2 y}{\cot y}\right) \frac{d y}{d x}+\log (\cot y)$
${\left[\log (\cot x)+x\left(\frac{\csc ^2 y}{\cot y}\right)\right] \frac{d y}{d x}=\log (\cot y)+y \frac{\csc ^2 x}{\cot _x}}$
$\frac{d y}{d x}=\frac{\log (\cot y)+y\left(\frac{\csc ^2 x}{\cot x}\right)}{\log (\cot x)+x\left(\frac{\csc ^2 y}{\cot y}\right)}$
$\frac{dy}{dx}= \frac{\log (\cot y)+y\left(\frac{1}{\sin x \cos x}\right)}{\log (\cot x)+x\left(\frac{1}{\sin y \cos y}\right)}$