If y x = e ? y - x , then d y d x is equal to:

Question

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

Smart Achievers · Accepted Answer

(4)

Ans. $\frac{{(1 + l o g y)}^{2}}{l o g y}$

Explanation:

We have, $y^{x} = e^{y - x}$

Taking log both sides, we get

$x \log y = y - x$

$\Rightarrow$ $x (\log y + 1) = y$

$\Rightarrow$ $\frac{y}{1 + \log y} = x$

On differentiating with respect to $x$ , we get

$(\frac{(1 + \log y) - (\frac{1}{y}) y}{{(1 + \log y)}^{2}}) \frac{d y}{d x} = 1$

$\Rightarrow$ $\frac{\log y}{{(1 + \log y)}^{2}} \cdot \frac{d y}{d x} = 1$

$\Rightarrow$ $\frac{d y}{d x} = \frac{{(1 + \log y)}^{2}}{\log y}$

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