The equation of the normal to the curve y=x^-x at the point of its ma…

We have,

$y=x^{-x}=e^{-x \log _e x}, x>0$

$\Rightarrow \frac{d y}{d x}=x^{-x}\left(-1-\log _e x\right)$

$\Rightarrow \frac{d y}{d x}=-x^{-x}\left(1+\log _e x\right)$

For the point of local maximum, we must have

$\frac{d y}{d x}=0 \Rightarrow 1+\log _e x=0 \Rightarrow x=\frac{1}{e}$

Clearly, $1+\log _e x<0$ for $0<x<\frac{1}{e}$ and $1+\log _e x>0$ for $x>\frac{1}{e}$

Thus,

$\frac{d y}{d x}>0$ for $0<x<\frac{1}{e}$ and $\frac{d y}{d x}<0$ for $x>\frac{1}{e}$

$\Rightarrow x=\frac{1}{e}$ is the point of local maximum.

Clearly, $\frac{d y}{d x}=0$ at $x=\frac{1}{e}$. So, the normal at $x=\frac{1}{e}$ is parallel to y-axis and its equation is given by $x=\frac{1}{e}$.