Question

Find $\frac{dy}{dx}$ while:
$x^y+y^x=a^b$

Answer 1

Let $u=x^y$ and $v=y^x$

$\Rightarrow u+v=a^b$

$\Rightarrow \frac{du}{dx}+\frac{dv}{dx}=\frac{d}{dx}\left(a^b\right)=0$ since differential of a constant is $0$

$\Rightarrow \frac{du}{dx}+\frac{dv}{dx}=0$

$u=x^y\Rightarrow \log u=y\log x$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=y\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(y\right)$

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

$\Rightarrow \frac{du}{dx}=u(\frac{y}{x}+\log x\frac{dy}{dx})$

$v=y^x\Rightarrow \log v=x\log y$

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right)$

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

$\Rightarrow \frac{dv}{dx}=v(\frac{x}{y}\frac{dy}{dx}+\log y)$

$\therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=u(\frac{y}{x}+\log x\frac{dy}{dx})+v(\frac{x}{y}\frac{dy}{dx}+\log y)$

$\Rightarrow \frac{dy}{dx}=u\frac{y}{x}+u\log x\frac{dy}{dx}+v\frac{x}{y}\frac{dy}{dx}+v\log y$

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

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

$\therefore \frac{dy}{dx}=\frac{x^y\frac{y}{x}+y^x\log y}{1-x^y\log x-y^x\frac{x}{y}}$ where $u=x^y$ and $v=y^x$

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