Question

Find $\frac{dy}{dx}$ of the function

$x^y+y^x=1$

Answer 1

The given function is $x^y+y^x=1$

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

Then, the function becomes $u+v=1$

$\frac{du}{dx}+\frac{dv}{dx}=0$ ...(1)

$u=x^y$

$\Rightarrow \log u=\log \left(x^y\right)$

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

Differentiating both sides with respect to $x$, we obtain

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

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

$v=y^x$

$\Rightarrow \log v=\log \left(y^x\right)$

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

Differentiating both sides with respect to $x$, we obtain

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

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

$\Rightarrow \frac{dv}{dx}=y^x(\log y+\frac{x}{y}.\frac{dy}{dx})$ ...(3)

From (1), (2) and (3) we obtain

$x^y(\log x\frac{dy}{dx}+\frac{y}{x})+y^x(\log y+\frac{x}{y}.\frac{dy}{dx})=0$

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

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