Question

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

Answer 1

Let $u=y^x,v=x^y,w=x^x$
$\left(i\right)u=y^x\Rightarrow logu=xlogy\Rightarrow \frac{du}{dx}=y^x[logy+\frac{x}{y}\frac{dy}{dx}]$
$\left(ii\right)v=x^y\Rightarrow logv=ylogx\Rightarrow \frac{dv}{dx}=x^y[logx\frac{dy}{dx}+\frac{y}{x}]$
$\left(iii\right)w=x^x\Rightarrow logw=xlogx\Rightarrow \frac{dw}{dx}=x^x[logx+1]$
$Now{y}^x+x^y+x^x=a^b\Rightarrow u+v+w=a^b\Rightarrow \frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0$
$\Rightarrow y^x[logy+\frac{x}{y}\frac{dy}{dx}]+x^y[logx\frac{dy}{dx}+\frac{y}{x}]+x^x[logx+1]=0$
$\therefore \frac{dy}{dx}=-\frac{y^{x}logy+y{x}^{y-1}+x^x[logx+1]}{logx+x{y}^{x-1}}=0$