Question

Find $\frac{dy}{dx}$of the functions given in question.

$x^y+y^x=1$

Answer 1

Given $x^y+y^x=1$

Let $u=x^y,v=y^x$

$\therefore$ u + v = 1

Differentiating w.r.t. x, $\frac{du}{dx}+\frac{dv}{dx}=0$

$\frac{du}{dx}+\frac{dv}{dx}=0$

Now, $u=x^y$

Taking log on both sides, we get log u = y log x

Differentiating w.r.t. x

$\Rightarrow \frac{1}{u}\frac{du}{dx}=y\times \frac{1}{x}+logx\frac{dy}{dx}\left(Usingproductrule\right)\Rightarrow \frac{du}{dx}=u[\frac{y}{x}+logx\frac{dy}{x}]\Rightarrow \frac{du}{dx}=x^y[\frac{y}{x}+logx\frac{dy}{dx}]\Rightarrow \frac{du}{dx}=y{x}^{y-1}+x^y.logx\frac{dy}{dx}Now,v=y^{x}Takinglogonbothsides\Rightarrow logv=xlogyDifferentiatingw.r.t.x,\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\times \frac{1}{y}\frac{dy}{dx}+logy\Rightarrow \frac{dv}{dx}v[\frac{x}{y}\frac{dy}{dx}+logy]\Rightarrow \frac{dv}{dx}{y}^x[\frac{x}{y}\frac{dy}{dx}+logy]\frac{dv}{dx}=x{y}^{x-1}\frac{dy}{dx}+y^{x}logyNow,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.\left(i\right).y{x}^{y-1}+x^y.logx\frac{dy}{dx}=x{y}^{x-1}\frac{dy}{dx}+y^{x}logy=0\Rightarrow \frac{dy}{dx}[x^{y}logx+x{y}^{x-1}]=-y^{x}logy+y{x}^{y-1}\Rightarrow \frac{dy}{dx}=-\frac{y^{x}logy+y{x}^{y-1}}{x^{y}logx+x{y}^{x-1}}$