Find dy dx of the functions given in Exercise: ( )^y=( )^x - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$

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Answer

Given: $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}\ \Rightarrow\ \log(\cos\text{x})^\text{y}=\log(\cos\text{y})^\text{x}$
$\Rightarrow\ \text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}\ \Rightarrow\ \frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{y})$
$\Rightarrow\ \text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{y}+\log\cos\text{y}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\frac{\text{d}}{\text{dx}}\cos\text{y}+\log\cos\text{y}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\Big(-\sin\text{y}\frac{\text{dy}}{\text{dx}}\Big)+\log\cos\text{y}$
$\Rightarrow\ -\text{y}\tan\text{x}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=-\text{x}\tan\text{y}.\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\ \text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}(\text{x}\tan\text{y}+\log\cos\text{x})=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\tan\text{x}+\log\cos\text{y}}{\text{x}\tan\text{y}+\log\cos\text{x}}$

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