If x^y = e^ x-y then dy dx is: - Vidyadip

MCQ

If $x^y = e^{x-y}$ then $\frac{\text{dy}}{\text{dx}}$ is:

A
$\frac{1+\text{x}}{1+\log\text{x}}$
B
$\frac{1-\log\text{x}}{1+\log\text{x}}$
C
$\text{Not defined.}$
✓
$\frac{\log\text{x}}{(1+\log\text{x})^2}$

✓

Answer

Correct option: D.

$\frac{\log\text{x}}{(1+\log\text{x})^2}$

We have, $x^y = e^{x-y}$
Taking $\log$ on both sides we get,
$\Rightarrow\text{y}\log\text{x}=(\text{x}-\text{y})\log)_\text{e}\text{e}$
$\Rightarrow\text{y}\log\text{x}=\text{x}-\text{y}$
$\Rightarrow\text{y}\log\text{x}+\text{y}=\text{x}$
$\Rightarrow\text{y}(1+\log\text{x})=\text{x}$
$\Rightarrow\text{y}=\frac{\text{x}}{(1+\log\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{x})\times1-\text{x}\times\Big(1+\frac{1}{\text{x}}\Big)}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\log\text{x}-1}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{(1+\log\text{x})^2}$

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