## lovesit2x Group Title Find (d^(2)y)/(dx^(2)) for the following function. y=e^(-4x^(2)) one year ago one year ago

1. abb0t Group Title

So find the second derivative of the function.

2. abb0t Group Title

This requires that you use the chain rule.

3. lovesit2x Group Title

$Find \frac{ d ^{2}y }{ dx ^{2} } for the following function. y=e ^{-4x ^{2}}$

4. abb0t Group Title

First, start by using chain rule. For the second derivative, you will be using product rule AND chain rule.

5. Xavier Group Title

You want $\frac{d}{dx}\left( \frac{dy}{dx}\right)$

6. satellite73 Group Title

first derivative is $-8xe^{-4x^2}$ by the chain rule second derivative requires the product rule

7. zepdrix Group Title

Do you understand how Sat got that first derivative? c:

8. abb0t Group Title

Yes. I do.

9. lovesit2x Group Title

yes

10. zepdrix Group Title

$\large y'=-8xe^{-4x^2}$ So now you need to find the second derivative as was stated above. $$\large y''$$ To differentiate this a second time, you'll need to apply the Product Rule, very similarly to the way we did it in the last problem! :)

11. zepdrix Group Title

Fine fine fine D: I'll do the pretty colors like last time. Maybe that will help.

12. lovesit2x Group Title

lol

13. zepdrix Group Title

$\huge y''=\color{royalblue}{\left(-8x\right)'}e^{-4x^2}-8x\color{royalblue}{\left(e^{-4x^2}\right)'}$Product Rule setup for the second derivative. The second blue term should give you the same thing you got for the first derivative!

14. lovesit2x Group Title

omg, its that easy??? why am i overthinking this

15. zepdrix Group Title

:3

16. zepdrix Group Title

What do you get for the first set of blue brackets?

17. lovesit2x Group Title

-8

18. lovesit2x Group Title

n the second is the -8x e ^-4x^2

19. zepdrix Group Title

$\huge y''=\color{orangered}{\left(-8\right)}e^{-4x^2}-8x\color{orangered}{\left(-8xe^{-4x^2}\right)}$ Like that? Yay good job!

20. lovesit2x Group Title

i love that lol