Question

If $x^y=a^x$, prove that $\frac{dy}{dx}=\frac{x{\log }_{e}a-y}{x{\log }_{e}x}$.

Answer 1

Given :

$x^y=a^x$

Taking log on both sides we get

$\log \left(x^y\right)=\log \left(a^x\right)$

$y\log x=x\log a$

Differentiating both sides we get

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

$⟹\log x\frac{dy}{dx}=\log a-\frac{y}{x}$

$⟹{\log }_{e}x\frac{dy}{dx}=\frac{x{\log }_{e}a-y}{x}$

$⟹\frac{dy}{dx}=\frac{x{\log }_{e}a-y}{x{\log }_{e}a}$