Question

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

Answer 1

Here, $x^y+y^x=a^b$
$\text{Let}u+v=a^b\text{where}u=x^y\text{and}v=y^x\therefore \frac{du}{dx}+\frac{dv}{dx}=0...\left(i\right)$

Now,
$u=x^y\Rightarrow \log u=y\log x\Rightarrow \frac{1}{u}\frac{du}{dx}=y.\frac{1}{x}+\log x.\frac{dy}{dx}\Rightarrow \frac{du}{dx}=x^y(\frac{y}{x}+\log x\frac{dy}{dx})$

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

Now,$x^y(\frac{y}{x}+\log x\frac{dy}{dx})+y^x(\frac{x}{y}\frac{dy}{dx}+\log y)=0\left[\text{using}\right(i\left)\right]\Rightarrow (x^y\log x+x{y}^{x-1})\frac{dy}{dx}=-(y^x\log y+y{x}^{y-1})\Rightarrow \frac{dy}{dx}=-\frac{y^x\log y+y{x}^{y-1}}{x^y\log x+x{y}^{x-1}}$