Question

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

Answer 1

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

Answer 2

Solution:

Given

$$x^y = e^{x-y}.$$

Taking natural logarithms on both sides,

$$\ln(x^y) = \ln\left(e^{x-y}\right) \implies y\ln x = x - y.$$

Rearrange:

$$y\ln x + y = x \implies y(1 + \ln x) = x \implies y = \frac{x}{1+\ln x}.$$

Differentiate using the quotient rule with $u = x$ and $v = 1+\ln x$ (so that $u'=1$ and $v'=\frac{1}{x}$):

$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{1\cdot (1+\ln x) - x\cdot \frac{1}{x}}{(1+\ln x)^2} = \frac{1+\ln x - 1}{(1+\ln x)^2} = \frac{\ln x}{(1+\ln x)^2}.$$

Explanation (Core):

1. Take $\ln$ on both sides: $y\ln x = x-y$.
2. Solve for $y$: $y = \frac{x}{1+\ln x}$.
3. Differentiate using the quotient rule to get $\displaystyle \frac{dy}{dx} = \frac{\ln x}{(1+\ln x)^2}$.