Question

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

Answer 1

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

$x\log y=y-x$ --- (1)

Differentiate with respect to $x$,

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

From equation (1), $\frac{x}{y}=\frac{1}{1+\log y}$

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

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

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

$\Rightarrow \frac{dy}{dx}=\frac{(1+\log y{)}^2}{\log y}$

Hence proved.