# Making this derivative look prettier

## Homework Statement

Find the 1st and 2nd derivative:

$$f(x) = x^{1/3}* e^{-x^2}$$

## The Attempt at a Solution

$$f'(x) = x^{1/3} * -2xe^{-x^2} + e^{-x^2} * \frac {1}{3}x^{-2/3}$$

I simplified this to:
$$[e^{-x^2}]*[-2x^{4/3} + \frac {1}{3}x^{-2/3}]$$

Also to find the x values is

$$-2x^{4/3} + \frac {1}{3}x^{-2/3}$$
$$x^{-2/3}*[-2x^{2} + \frac {1}{3}]$$

$$x^{2} = 1/6$$
$$x = +- \sqrt{1/6}$$

Okay now I'm trying to make the 2nd derivative be simplified so I can solve for the inflection points.

$$f''(x) = [e^{-x^{2}}]*[\frac {-8}{3}x^{1/3} - \frac {2}{9}x^{-5/3}] + [-2x^{4/3} + \frac {1}{3}x^{-2/3}]*[-2xe^{-x^{2}}]$$

Im kind of stuck after this. I think it would involve taking out an $$e^{-x{2}}$$ but it's very confusing.

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tiny-tim
Homework Helper
… just keep going …

Hi Flappy!

Just keep going … you should get $$}x^{-5/3} e^{-x^2}$$ times a quadratic in x^2.

Hmm I tried to do this:

$$e^{-x^{2}}[ -\frac {8}{3}x^{1/3} - \frac {2}{9}x^{-5/3} + 4x^{7/3} - \frac {2}{3}x^{1/3}]$$

I multiplied the -2x and then took out the e^(-x^2). Would this be right? I'm not sure where you're getting x^(-5/3)

Ah, i think i see it. There's a common factor of x^(-5/3)