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

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

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

$\Rightarrow$ $x\log y = y - x \cdot {\log _e} = (y - x)$

$\Rightarrow$ $\log y = \frac{{(y - x)}}{x}$ ……..(i)

Now, differentiating w.r.t. $x$, we get
$\frac{d}{{dy}}\log y \cdot \frac{{dy}}{{dx}} = \frac{d}{{dx}}\frac{{(y - x)}}{x}$

$\Rightarrow$ $\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \frac{{x \cdot \frac{d}{{dx}}(y - x) - (y - x) \cdot \frac{d}{{dx}} \cdot x}}{{{x^2}}}$

$\Rightarrow$ $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{x\left( {\frac{{dy}}{{dx}} - 1} \right) - (y - x)}}{{{x^2}}}$

$\Rightarrow$ $\frac{{{x^2}}}{y} \cdot \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} - x - y + x$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left( {\frac{{{x^2}}}{y} - x} \right) = - y$

therefore,$\frac{{dy}}{{dx}} = \frac{{ - {y^2}}}{{{x^2} - xy}} = \frac{{ - {y^2}}}{{x(x - y)}}$

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

Hence proved.