# What is the equation of the line normal to  f(x)=x^2/e^x-x/e^(x^2)  at  x=0?

Dec 29, 2015

$y = x$

#### Explanation:

$\textcolor{b l u e}{\text{Graph of } f \left(x\right)}$
graph{(x^2)/(e^x)-x/(e^(x^2)}

$f \left(0\right) = 0$

Since the Normal passes through $\left(0 , 0\right)$, it has the form of $y = m x$.

Since the Normal is perpendicular to the tangent of $f$ at $x = 0$, $m \cdot f ' \left(0\right) = - 1$.

To find $f ' \left(0\right)$, we first find the general $f ' \left(x\right)$, and then substitute $x = 0$.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{{x}^{2}}{{e}^{x}} - \frac{x}{{e}^{{x}^{2}}}\right)$

$= \frac{{e}^{x} \left(2 x\right) - {x}^{2} \left({e}^{x}\right)}{{\left({e}^{x}\right)}^{2}} - \frac{{e}^{{x}^{2}} \left(1\right) - x \left(2 x {e}^{{x}^{2}}\right)}{{\left({e}^{{x}^{2}}\right)}^{2}}$

$= \frac{2 x - {x}^{2}}{{e}^{x}} - \frac{1 - 2 {x}^{2}}{{e}^{{x}^{2}}}$

$f ' \left(0\right) = - 1$

Therefore, $m = \frac{- 1}{- 1} = 1$

Equation of Normal: $y = x$

$\textcolor{b l u e}{\text{Graph of Normal}}$
graph{x}