# How do you find the 1st and 2nd derivative of e^(x^2)?

Jul 16, 2016

$f ' \left(x\right) = 2 x {e}^{{x}^{2}} , f ' ' \left(x\right) = 2 {e}^{{x}^{2}} \left(2 {x}^{2} + 1\right)$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x} \text{ and } \frac{d}{\mathrm{dx}} \left({e}^{g \left(x\right)}\right) = {e}^{g \left(x\right)} . g ' \left(x\right)$

$\textcolor{b l u e}{\text{First derivative}}$

$f \left(x\right) = {e}^{{x}^{2}} \Rightarrow f ' \left(x\right) = {e}^{{x}^{2}} .2 x = 2 x {e}^{{x}^{2}}$

$\textcolor{b l u e}{\text{Second derivative}}$

Differentiate using the $\textcolor{red}{\text{product rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = g \left(x\right) h \left(x\right) \Rightarrow f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Differentiating $f ' \left(x\right) = 2 x {e}^{{x}^{2}}$

now $g \left(x\right) = 2 x \Rightarrow g ' \left(x\right) = 2$

and $h \left(x\right) = {e}^{{x}^{2}} \Rightarrow h ' \left(x\right) = 2 x {e}^{{x}^{2}}$

$\Rightarrow f ' ' \left(x\right) = 2 x .2 x {e}^{{x}^{2}} + 2 {e}^{{x}^{2}} = 4 {x}^{2} {e}^{{x}^{2}} + 2 {e}^{{x}^{2}}$

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