Question

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

Answer 1

Given, $x^x+x^y+y^x=a^b$
Let $p=x^x,q=x^y,r=y^x$
Let us take $p=x^x$
Take $\log$ on both sides
$\log p=x\log x$
$\Rightarrow \frac{1}{p}\frac{dp}{dx}=\log x+\frac{x}{x}=\log x+1=\log x+\log e$
$\Rightarrow \frac{dp}{dx}=x^x\log ex$ ....(1)
Now lets take $q=x^y$
Take $\log$ on both sides
$\log q=y\log x$
$\Rightarrow \frac{1}{q}\frac{dq}{dx}=\frac{y}{x}+\log x\frac{dy}{dx}$
$\Rightarrow \frac{dq}{dx}=y{x}^{y-1}+\left(x^y\log x\right)\frac{dy}{dx}$ ....(2)
and $r=y^x$
Take $\log$ on both sides
$\log r=x\log y$
$\Rightarrow \frac{1}{r}\frac{dr}{dx}=\frac{x}{y}\frac{dy}{dx}+\log y$
$\Rightarrow \frac{dr}{dx}=y^x\log y+x{y}^{x-1}\frac{dy}{dx}$ ....(3)
$x^x+x^y+y^x=a^b⟹p+q+r=a^b$
Differentiate both sides with respect to $x$
$\frac{dp}{dx}+\frac{dq}{dx}+\frac{dr}{dx}=0$
$x^x\log ex+y{x}^{y-1}+y^x\log y+\frac{dy}{dx}[x^y\log x+x{y}^{x-1}]=0$
$\frac{dy}{dx}=-\left[\frac{x^x\log ex+y{x}^{y-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}\right]$