Question

$\text{If}{x}^y-y^x=a^b,$ find $\frac{dy}{dx}.$

Answer 1

$x^y-y^x=a^b$
Putting $u=x^y$ and $v=y^x$, we get
$u-v=a^b$
Differentiating with respect to $x$, we get
$\frac{du}{dx}-\frac{dv}{dx}=0\cdots \left(1\right)$
Now, $u=x^y$
Taking logarithm on both sides, we get
$\log u=y\log x$
Differentiating with respect to $x$, we get
$\frac{1}{u}\frac{du}{dx}=y\times \frac{1}{x}+\log x\times \frac{dy}{dx}$
$\Rightarrow \frac{du}{dx}=u[\frac{y}{x}+\log x\times \frac{dy}{dx}]$
$\Rightarrow \frac{du}{dx}=x^y[\frac{y}{x}+\log x\times \frac{dy}{dx}]$

Also, $v=y^x$
Taking logarithm on both sides, we get
$\log v=x\log y$
Differentiating with respect to $x$, we get
$\frac{1}{v}\frac{dv}{dx}=x\times \frac{1}{y}\frac{dy}{dx}+\log y\times 1$
$\Rightarrow \frac{dv}{dx}=v[\frac{x}{y}\frac{dy}{dx}+\log y]$
$\Rightarrow \frac{dv}{dx}=y^x[\frac{x}{y}\frac{dy}{dx}+\log y]$

From $\left(1\right),$ we get
$x^y[\frac{y}{x}+\log x\times \frac{dy}{dx}]-y^x[\frac{x}{y}\frac{dy}{dx}+\log y]=0$
$\Rightarrow [x^y\log x-x{y}^{x-1}]\frac{dy}{dx}=y^x\log y-y\times x^{y-1}$
$\Rightarrow \frac{dy}{dx}=\frac{y^x\log y-y\times x^{y-1}}{x^y\log x-x{y}^{x-1}}$