If x^y = e^{x - y},then frac{dy}{dx} at x = 1 | vedclass

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(C) Given equation is $x^y = e^{x - y}$.Taking $\log$ on both sides,we get $y \log x = (x - y) \log e = x - y$ . . . . . . $(i)$.When $x = 1$,substitu

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(C) Given equation is $x^y = e^{x - y}$.Taking $\log$ on both sides,we get $y \log x = (x - y) \log e = x - y$ . . . . . . $(i)$.When $x = 1$,substituting in $(i)$ gives $y \log 1 = 1 - y$,which implies $0 = 1 - y$,so $y = 1$.Differentiating both sides of $(i)$ with respect to $x$ using the product rule:$y \cdot (\frac{1}{x}) + \log x \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx}$.Rearranging the terms to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} (\log x + 1) = 1 - \frac{y}{x}$.$\frac{dy}{dx} = \frac{x - y}{x (\log x + 1)}$.At $x = 1$ and $y = 1$:$\frac{dy}{dx} = \frac{1 - 1}{1 (\log 1 + 1)} = \frac{0}{1} = 0$.

If $x^y = e^{x - y}$,then $\frac{dy}{dx}$ at $x = 1$ is . . . . . .

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