If x^y = e^ x –y , show that dy dx = log x log(x e) ^ 2 . - Vidyadip

Question

If $x^y = e^{x –y},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{log x}}{\left\{\text{log(x e)}\right\}^{2}}.$

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Answer

$x^y = e^{x–y} \Rightarrow y . \log x = (x – y) \log e = x – y$
$\text{y}=\frac{\text{x}}{\text{1 + log x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{(1 + log x)}\cdot\text{1 - x}\cdot\Big(\frac{1}{\text{x}}\Big)}{\text{(1 + log x)}^{2}}=\frac{\text{log x}}{\text{(1 + log x)}^{2}}$
$=\frac{\log\text{x}}{\text{(log e + log x)}^{2}}=\frac{\text{log x}}{\text{[log(xe)]}^{2}}$.

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