Question
If $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$

Answer

We have, $(\sin\text{x})^{\text{y}}=(\cos\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\sin\text{x})^\text{y}=\log(\cos\text{y})^{\text{x}}$
$\Rightarrow\text{y}\log(\sin\text{x})=\text{x}\log(\cos\text{y})$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\big[\text{y}\log\sin\text{x}\big]=\frac{\text{d}}{\text{dx}}\big[\text{x}\log\cos\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\text{x}\frac{\text{dy}}{\text{dx}}(\log\cos\text{y})+\log\cos\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\text{y}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}\frac{\text{x}}{\text{dx}}(\cos\text{y})+\log\cos\text{y}(1)$
$\Rightarrow\frac{\text{y}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =\frac{\text{x}}{\cos\text{y}}(-\sin\text{y})\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow \text{y}\cot\text{x}+\log\sin\text{x}\frac{\text{dy}}{\text{dx}} \\ =-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\sin\text{x}+\text{x}\tan\text{y}) \\ =\log\cos\text{y}-\text{y}\cot\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}-\text{y}\cot\text{x}}{\log\sin\text{x}+\text{x}\tan\text{y}}$

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