# Thread: funky derivative and second of e

1. ## funky derivative and second of e

I am doing working a derivative of E
Can you tell me if I am doing it right.

$y=-e^{\frac{-x^2}{2}$

chain rule:
$f(x)= -e^u$
$f'= -e^u$
$g(x)=\frac{-x^2}{2}$
$g'=-x$

$=-e^{\frac{-x^2}{2}}*-x$
SO, Y'= $xe^{\frac{-x^2}{2}$

for the second derivative, Y'', I use product rule, I believe.
product rule
$(1)*(e^{\frac{-x^2}{2}})+(x)*(-xe^{\frac{-x^2}{2}})$
$Y''=e^{\frac{-x^2}{2}}-x^2e^{\frac{-x^2}{2}$

Did I do it right?

2. ## Re: funky derivative and second of e

Looks good to me.

3. ## Re: funky derivative and second of e

Thanks you for replying. Perhaps you can further help me with this problem. The answer continues to evade me.I am trying to find the critical points of the equation. The book says to find the critical points, set f' = 0 .

According to mathlab, I should get critical points at x=-1 and x=1.

Unfortunately, I have been trying, unsuccessfully for hours, to see how Y' or
$xe^{\frac{-x^2}{2}$ = 0 gives me a X=-1 and X=1

Can you help me?

4. ## Re: funky derivative and second of e

Originally Posted by delgeezee
Thanks you for replying. Perhaps you can further help me with this problem. The answer continues to evade me.I am trying to find the critical points of the equation. The book says to find the critical points, set f' = 0 .

According to mathlab, I should get critical points at x=-1 and x=1.

Unfortunately, I have been trying, unsuccessfully for hours, to see how Y' or
$xe^{\frac{-x^2}{2}$ = 0 gives me a X=-1 and X=1

Can you help me?
The only critical point of y is x = 0, found by solving dy/dx = 0.

It is dy/dx that has critical points at x = 1 and x = -1, found by solving d^2y/dx^2 = 0.

You need to think more carefully about what you're trying to do and why.

5. ## Re: funky derivative and second of e

I got it now. I was looking for inflection points and not all critical points are inflection points. I also made a better routine for solving so I can stay organized as I go through the problem.