Making this derivative look prettier

[Flappy]

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.

 

[tiny-tim]

… just keep going …

Hi Flappy!

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

 

[Flappy]

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)

 

[Flappy]

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