# What is the second derivative of f(x) = e^(-x^2 ?

Apr 8, 2016

Answer: $f ' ' \left(x\right) = - 2 {e}^{- {x}^{2}} + {\left(2 x\right)}^{2} {e}^{{x}^{2}} = 2 {e}^{{x}^{2}} \left[{x}^{2} - 1\right]$

#### Explanation:

Given: f(x) = e^(-x^2

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

Definitions and Principles:
a) Derivative of - $f ' \left(x\right) = \frac{{\mathrm{de}}^{x}}{\mathrm{dx}} = {e}^{x}$

b) Chain Rule - $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(x\right)}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

c) Product Rule - $\frac{d \left[f \left(x\right) \cdot r \left(x\right)\right]}{\mathrm{dx}} = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} \cdot r \left(x\right) + \frac{\mathrm{dr} \left(x\right)}{\mathrm{dx}} \cdot f \left(x\right)$

Solution Strategy:
1) Set up $\frac{{d}^{2} f \left(x\right)}{\mathrm{dx}} ^ 2$ using b) the chain rule, use a)
2) For 2nd derivative you will have to set up chain and product rule

Solution:
1) Let $u = - {x}^{2}$ then we we can $g \left(u\right) = {e}^{u}$
$h \left(x\right) = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{dg} \left(u\right)}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$h \left(x\right) = \frac{{\mathrm{de}}^{u}}{\mathrm{du}} \cdot \frac{{\mathrm{dx}}^{2}}{\mathrm{dx}} = {e}^{u} \left(- 2 x\right) {|}_{u = {x}^{2}} = - 2 x {e}^{{x}^{2}}$

2) Now we need to take the derivative of h(x).
Let r(x)=2x and f(x)=-e^(-x^2
$h ' \left(x\right) = \frac{\mathrm{dr} \left(x\right)}{\mathrm{dx}} \cdot f \left(x\right) + \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} \cdot r \left(x\right)$
$h ' \left(x\right) = - 2 {e}^{- {x}^{2}} - 2 x \cdot \textcolor{red}{\frac{{\mathrm{de}}^{- {x}^{2}}}{\mathrm{dx}}}$

Now from 1) we know:

$f ' \left(x\right) = \textcolor{red}{\frac{{\mathrm{de}}^{- {x}^{2}}}{\mathrm{dx}} = - 2 x {e}^{{x}^{2}}}$
$h ' \left(x\right) = - 2 {e}^{- {x}^{2}} - \left(2 x\right) \cdot \textcolor{red}{- 2 x {e}^{{x}^{2}}}$

Answer: $f ' ' \left(x\right) = - 2 {e}^{- {x}^{2}} + {\left(2 x\right)}^{2} {e}^{{x}^{2}} = 2 {e}^{{x}^{2}} \left[{x}^{2} - 1\right]$