Question

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

Answer 1

$x^x+x^y+y^x=ab...........\left(i\right)Letu=x^{x}logu=xlogx\frac{1}{u}.\frac{du}{dx}=x.\frac{1}{x}+logx\therefore \frac{du}{dx}=x^x(1+logx)Letv=x^{y}logv=ylogx\frac{1}{v}.\frac{dv}{dx}=(\frac{y}{x}+logx\frac{dy}{dx})\therefore \frac{dv}{dx}=x^y(\frac{y}{x}+logx\frac{dy}{dx})$

$Letw=y^{x}Logw=xlogy\frac{1}{w}.\frac{dw}{dx}=(\frac{x}{y}\frac{dy}{dx}+logy)$

$\therefore \frac{dw}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+logy)$

(i) can be written as

$u+v+w=a^b\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0\Rightarrow x^x(1+logx)+x^y(\frac{y}{x}+logx\frac{dy}{dx})+y^x(\frac{x}{y}\frac{dy}{dx}+logy)=0\Rightarrow x^x+x^{x}logx+x^y.\frac{y}{x}+x^y.logx.\frac{dy}{dx}+y^x.\frac{x}{y}\frac{dy}{dx}+y^{x}logy=0\Rightarrow \frac{dy}{dx}(x^y.logx+y^x.\frac{x}{y})=x^x+x^{x}logx+x^y.\frac{y}{x}+y^{x}logy\Rightarrow \frac{dy}{dx}(x^y.logx+x{y}^{x-1})=(x^x+x^{x}logx+)\therefore \frac{dy}{dx}\frac{(x^x+x^{x}logx+y{x}^{y-1}+y^{x}logy)}{(x^y.logx+x{y}^{x-1}).}$