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

Question

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

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Answer

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

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