Question

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

Answer 1

We have , $y^x=e^{y-x}$

Taking log both side

$\Rightarrow \log y^x=\log e^{y-x}$

$\Rightarrow x\log y=y-x.......\left(1\right)$

Differentiating w.r.t x, we get

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

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

$=\frac{y(\log y+1)}{y-\frac{y}{1+\log y}}$ [Using (1)]

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