Question

Find $\frac{dy}{dx}$ of function
$x^y+y^x=1$

Answer 1

The correct answer is $∴\frac{dy}{dx}=\frac{-yx^{y-1}+y^xlogy}{x^ylogx+xy^{x-1}}$
The given function is $x^y+y^x=1$
Let $x^y =u$ and $y^x=v$
Then, the function becomes $u+v=1$
$∴\frac{du}{dx}+\frac{dv}{dx}=0 ....(1)$
$u=x^y$
$⇒log\,u=log(x^y)$
$⇒log\,u=ylogx$
Differentiating both sides with respect to $x$,we obtain
$⇒\frac{1}{u}\frac{du}{dx}=logx\frac{dy}{dx}+y.\frac{d}{dx}(logx)$
$⇒\frac{du}{dx}=u[logx\frac{dy}{dx}+y.\frac{1}{x}]$
$⇒\frac{du}{dx}=x^y[logx\frac{dy}{dx}+\frac{y}{x}] ..(2)$
$v=y^x$
$⇒log\,v=log(y^x)$
$⇒log\,v=xlogy$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=logy.\frac{d}{dx}(x)+x.\frac{d}{dx}(logy)$
$⇒\frac{dv}{dx}=v(logy.1+x.\frac{1}{y}.\frac{dy}{dx})$
$⇒\frac{dv}{dx}=y^x(logy+\frac{x}{y}.\frac{dy}{dx}) ...(3)$
Therefore,from (1),(2),and(3),we obtain
$x^y[logx\frac{dy}{dx}+\frac{y}{x}]+y^x(logy+\frac{x}{y}.\frac{dy}{dx})=0$
$⇒(x^ylogx+xy^{x-1})\frac{dy}{dx}=-(yx^{y-1}+y^xlogy)$
$∴\frac{dy}{dx}=\frac{-yx^{y-1}+y^xlogy}{x^ylogx+xy^{x-1}}$