Find d y d x of the function : x^y+y^x=1

Question

Find $\frac{d y}{d x}$ of the function : $x^y+y^x=1$

✓

Answer

Suppose, $u=x^y$ and $v=y^x$
$\therefore u+v=1$
Now, differentiate w.r.t. $x$,
$\frac{d u}{d x}+\frac{d v}{d x}=0\ldots(1)$
Here, $u=x^y$
Take both the sides $\log$,
$\log u=y \log x$
Now, differentiate w.r.t. $x$,
$\begin{aligned}
\frac{d u}{d x} \frac{1}{u} & =y \frac{d}{d x} \log x+\log x \frac{d}{d x} y \\
\therefore \quad \frac{d u}{d x} \frac{1}{u} & =y \cdot \frac{1}{x}+\log x \frac{d y}{d x} \\
\therefore \quad \frac{d u}{d x} & =u\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \\
& =x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \\
\therefore \quad \frac{d u}{d x} & =x^{y-1} y+x^y \log \frac{d y}{d x}\ldots(2)
\end{aligned}$
Now, $v=y^x$
Take both the side $\log$,
$\log v=x \log y$
Now, differentiate w.r.t. $x$,
$\begin{aligned}
\frac{1}{v} \frac{d v}{d x} & =x \frac{d}{d x} \log y+\log y \frac{d}{d x} x \\
\therefore \quad \frac{1}{v} \frac{d v}{d x} & =x \cdot \frac{1}{y} \frac{d y}{d x}+\log y \\
\therefore \quad \frac{d v}{d x} & =v\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]
\end{aligned}$
$=y^x\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]\ldots(3)$
Put the value of equation (2) and (3) in equation (1),
$\begin{array}{l}
x^{y-1} y+x^y \log x \frac{d y}{d x}+y^{x-1} x \frac{d y}{d x}+y^x \log y=0 \\
\frac{d y}{d x}\left[x^y \log x+y^{x-1} x\right]=-y^x \log y-x^{y-1} y \\
\frac{d y}{d x}=-\frac{\left[y^x \log y+x^{y-1} y\right]}{\left[x^y \log x+y^{x-1} x\right]}
\end{array}$