How do you find the first and second derivative of y=e^(x^3)?

Dec 15, 2016

$\frac{d}{\mathrm{dx}} {e}^{{x}^{3}} = 3 {x}^{2} {e}^{{x}^{3}}$
${d}^{\left(2\right)} / \left({\mathrm{dx}}^{2}\right) {e}^{{x}^{3}} = 6 x {e}^{{x}^{3}} \left(1 + \frac{3}{2} {x}^{3}\right)$

Explanation:

Using the chain rule:

$\frac{d}{\mathrm{dx}} {e}^{{x}^{3}} = \frac{{\mathrm{de}}^{{x}^{3}}}{d \left({x}^{3}\right)} \cdot \frac{d \left({x}^{3}\right)}{\mathrm{dx}} = 3 {x}^{2} {e}^{{x}^{3}}$

Then using the product rule:

${d}^{\left(2\right)} / \left({\mathrm{dx}}^{2}\right) {e}^{{x}^{3}} = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} {e}^{{x}^{3}}\right) = \left(\frac{d}{\mathrm{dx}} 3 {x}^{2}\right) {e}^{{x}^{3}} + 3 {x}^{2} \left(\frac{d}{\mathrm{dx}} {e}^{{x}^{3}}\right) = 6 x {e}^{{x}^{3}} + 3 {x}^{2} \cdot 3 {x}^{2} {e}^{{x}^{3}} = 6 x {e}^{{x}^{3}} + 9 {x}^{4} {e}^{{x}^{3}} = 6 x {e}^{{x}^{3}} \left(1 + \frac{3}{2} {x}^{3}\right)$