First slide

Derivatives

Question

If $x^{y} = e^{x - y}$ ,then $\frac{dy}{dx}$ is equal to

Easy

A

$\frac{1 + x}{1 + \log x}$
B

$\frac{1 - \log x}{1 + \log x}$
C

not defined
D

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

Solution

Since, $x^{y} = e^{x - y}$
Taking log on both sides, we get
$\begin{aligned} ylog x = (x - y) \log_{e} e \\ \Rightarrow y = \frac{x}{1 + \log x} \end{aligned}$
$On differentiating w . r . t . x, we ge$ t
$\frac{dy}{dx} = \frac{(1 + \log x) - x \cdot \frac{1}{x}}{(1 + \log x)^{2}} = \frac{\log x}{(1 + \log x)^{2}}$

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