What is the second derivative of f(x)= xe^(x^3)?

Jan 16, 2016

$f ' ' \left(x\right) = 3 \left(3 {x}^{3} + 4\right) {x}^{2} {e}^{{x}^{3}}$

Explanation:

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

$f \left(x\right) = h \left(x\right) \cdot {e}^{g \left(x\right)}$

h(x)=x;

$g \left(x\right) = {x}^{3}$

We have to apply Product Rule and Chain Rule
Let's start founding $f ' \left(x\right)$

$f ' \left(x\right) = h ' \left(x\right) {e}^{g \left(x\right)} + h \left(x\right) \cdot \left[\frac{d}{\mathrm{dx}} {e}^{g} \left(x\right) \cdot \frac{d}{\mathrm{dx}} g \left(x\right)\right] =$

$= 1 \cdot {e}^{{x}^{3}} + x \cdot \left[{e}^{{x}^{3}} \cdot 3 {x}^{2}\right] = \left(1 + 3 {x}^{3}\right) {e}^{{x}^{3}}$

Now we can computate $f ' ' \left(x\right)$ applying again Product Rule and Chain Rule

$f ' ' \left(x\right) = \left[\frac{d}{\mathrm{dx}} \left(1 + 3 {x}^{3}\right)\right] {e}^{{x}^{3}} + \left(1 + 3 {x}^{3}\right) \cdot \left[\frac{d}{\mathrm{dx}} {e}^{{x}^{3}} \frac{d}{\mathrm{dx}} {x}^{3}\right] =$

$= \left[0 + 3 \cdot 3 {x}^{2}\right] {e}^{x} + \left(1 + 3 {x}^{3}\right) \left[{e}^{{x}^{3}} \cdot 3 {x}^{2}\right] =$

$= 3 \cdot 3 {x}^{2} {e}^{{x}^{3}} + \left(1 + 3 {x}^{3}\right) \cdot 3 {x}^{2} {e}^{{x}^{3}} =$

$= 3 {x}^{2} {e}^{{x}^{3}} \left(3 + 1 + 3 {x}^{3}\right) =$

$= 3 \left(3 {x}^{3} + 4\right) {x}^{2} {e}^{{x}^{3}}$