If x^y = e^(x-y), prove that (dy)/(dx) = (log x)/((1+log x)²)

If $x^y = e^{x-y}$, prove that $\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$

Given: $x^y = e^{x-y}$

To show: $\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$

Answer

Proof that $\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}$

1. Take logs of $x^y=e^{x-y}$: $y\log x=x-y$.
2. Collect $y$: $y(1+\log x)=x$, so $y=\frac{x}{1+\log x}$.
3. Differentiate by the quotient rule: $\frac{dy}{dx}=\frac{(1+\log x)\cdot1-x\cdot\frac1x}{(1+\log x)^2}$.
4. Simplify the numerator: $1+\log x-1=\log x$.
5. Hence $\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}$.