If y^x=e^ x-e , prove that dy dx = (1+ )^2 - Vidyadip

Question

If $\text{y}^\text{x}=\text{e}^{\text{x}-\text{e}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$

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Answer

We have, $\text{y}^\text{x}=\text{e}^{\text{y}-\text{x}},$
$\Rightarrow\ \log\text{y}^\text{x}=\log^{\text{y}-\text{x}}$
$\Rightarrow\ \text{x}\log\text{y}=\text{y}-\text{x}\log_\text{e}=(\text{y}-\text{x})$ $[\because\log_\text{e}=1]$
$\Rightarrow\ \log\text{y}=\frac{(\text{y}-\text{x})}{\text{x}}\ \ \dots(\text{i})$
Now, differentiating w.r.t. x, we get
$\frac{\text{d}}{\text{dy}}\log\text{y}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\frac{(\text{y}-\text{x})}{\text{x}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\cdot\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})-(\text{y}-\text{x})\cdot\frac{\text{d}}{\text{dx}}\text{x}}{\text{x}^2}$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)-(\text{y}-\text{x})}{\text{x}^2}$
$\Rightarrow\ \frac{\text{x}^2}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{x}-\text{y}+\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{y}}-\text{x}\Big)=-\text{y}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{\text{x}^2-\text{xy}}=\frac{-\text{y}^2}{\text{x}(\text{x}-\text{y})}$
$=\frac{\text{y}^2}{\text{x}(\text{y}-\text{x})}\cdot\frac{\text{x}}{\text{x}}=\frac{\text{y}^2}{\text{x}^2}\cdot\frac{1}{\frac{(\text{y}-\text{x})}{\text{x}}}$
$=\frac{(1+\log\text{y})^2}{\log\text{y}}$ $\Big[\because\log\text{y}=\frac{\text{y}-\text{x}}{\text{x}}\log\text{y}=\frac{\text{y}}{\text{x}}-1\Rightarrow1+\log\text{y}=\frac{\text{y}}{\text{x}}\Big]$
Hence proved.

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