Question

Let x>0, and y be a positive function of x such that x^y=y^-x.

Find y′

Answer 1

$x^{y}=y^{(-x)}$

Taking logerithm with respect to base e on both sides

$\ln{x^{y}}=\ln{y^{(-x)}}$

=> $y\ln{x}=-x\ln{y}$

Differentiating with respect to x

$\frac{d(y\ln{x})}{dx}= - \frac{d(x\ln{y})}{dx}$

=> $y\frac{d}{dx} (\ln{x})+\ln{x} \frac{dy}{dx}=-x\frac{d}{dx}(\ln{y})-\ln{y} \frac{d}{dx}{(x)}$

=> $y.\frac{1}{x} + ln{x} \frac{dy}{dx}=-x.\frac{1}{y}.\frac{dy}{dx}-\ln{y}$

=> $\ln{x} \frac{dy}{dx}+x.\frac{1}{y}.\frac{dy}{dx}$ = $-\frac{y}{x} - \ln{y}$

=> $x(y\ln{x}+x)\frac{dy}{dx}= -y(y+x\ln{y})$

=>

=> $\frac{dy}{dx}=$ $-\frac{ y(y+x\ln{y})}{x(y\ln{x}+x)}$

So , y' = $-\frac{ y(y+x\ln{y})}{x(y\ln{x}+x)}$