Question
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
✓
Answer
$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
Taking log on both sides we get
$\text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}-\text{y}\tan\text{x}=\log\cos\text{y}-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}+\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\cos\text{x}+\text{x}\tan\text{y})=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}+\text{y}\tan\text{x}}{\log\cos\text{x}+\text{x}\tan\text{y}}$
Taking log on both sides we get
$\text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}-\text{y}\tan\text{x}=\log\cos\text{y}-\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\log\cos\text{x}+\text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\log\cos\text{x}+\text{x}\tan\text{y})=\log\cos\text{y}+\text{y}\tan\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\cos\text{y}+\text{y}\tan\text{x}}{\log\cos\text{x}+\text{x}\tan\text{y}}$
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