Prove that: if $x^y = e^{x-y}$ then $\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}$

Question

$x^y = e^{x-y}$. Prove that $\dfrac{dy}{dx}=\dfrac{\log x}{(1+\log x)^2}$.

I have tried to solve the problem by taking log no both sides and then: $y \log x = x-y$. Then i differentiated both sides w.r.t $x$ and ended up with a term which has both $x$ and $y$.

Answer 1

$$y\log x=(x-y)$$ $$y(1+\log x)=x$$ $$y=\frac{x}{1+\log x}$$ we know that $$(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$ so $$y'=\frac{1(1+\log x)-x(\frac{1}{x})}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}$$

Answer 2

I don't see the problem. For one, note that $y = \dfrac{x}{1+\log x}$. $$\begin{equation} \begin{split} x^y = e^{x-y} & \implies y \log x = x-y \\ & \implies \frac{dy}{dx}\log x + \frac{y}{x} = 1-\frac{dy}{dx} \\ & \implies \frac{x-y}{x} = \frac{dy}{dx}(\log x+ 1) \\ & \implies \frac{dy}{dx} = \frac{x-y}{x\log x + x} \\ & \implies \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2} \end{split}\end{equation}$$

Yes, you will end up with a $y$ on the other side of the $\frac{dy}{dx}$. However, the value of $y$ can be found using the equation that $e^{x-y} = x^y$.

Answer 3

This should help! $y \log x =x - y \to y \log x + y = x \to y(\log x + 1) = x \to y = \frac{x}{\log x + 1}$