Question

If $e^{x-y}=x^y$, then prove that $\frac{dy}{dx}=\frac{\log x}{{\left[\log ex\right]}^2}$.

Answer 1

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

Applying ${\log }_e$ on both the sides

$\Rightarrow x-y=y\log x$ ...... $\left(i\right)$

Differentiating both the sides w.r.t x

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

$\Rightarrow (1+\log x)\frac{dy}{dx}=\frac{x-y}{x}$

$\Rightarrow (1+\log x)\frac{dy}{dx}=\frac{y\log x}{x}$

$\Rightarrow (1+\log x)\frac{dy}{dx}=\frac{\log x}{1+\log x}$ ...... Using $\left(i\right)$

$\Rightarrow (\log e+\log x)\frac{dy}{dx}=\frac{\log x}{(\log e+\log x)}$

$\Rightarrow \frac{dy}{dx}=\frac{\log x}{{\left[\log ex\right]}^2}$