anonymous
  • anonymous
Find (d^(2)y)/(dx^(2)) for the following function. y=e^(-4x^(2))
Calculus1
schrodinger
  • schrodinger
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abb0t
  • abb0t
So find the second derivative of the function.
abb0t
  • abb0t
This requires that you use the chain rule.
anonymous
  • anonymous
\[Find \frac{ d ^{2}y }{ dx ^{2} } for the following function. y=e ^{-4x ^{2}}\]

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abb0t
  • abb0t
First, start by using chain rule. For the second derivative, you will be using product rule AND chain rule.
anonymous
  • anonymous
You want \[\frac{d}{dx}\left( \frac{dy}{dx}\right)\]
anonymous
  • anonymous
first derivative is \[-8xe^{-4x^2}\] by the chain rule second derivative requires the product rule
zepdrix
  • zepdrix
Do you understand how Sat got that first derivative? c:
abb0t
  • abb0t
Yes. I do.
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
\[\large y'=-8xe^{-4x^2}\] 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! :)
zepdrix
  • zepdrix
Fine fine fine D: I'll do the pretty colors like last time. Maybe that will help.
anonymous
  • anonymous
lol
zepdrix
  • zepdrix
\[\huge y''=\color{royalblue}{\left(-8x\right)'}e^{-4x^2}-8x\color{royalblue}{\left(e^{-4x^2}\right)'}\]Product Rule setup for the second derivative. The second blue term should give you the same thing you got for the first derivative!
anonymous
  • anonymous
omg, its that easy??? why am i overthinking this
zepdrix
  • zepdrix
:3
zepdrix
  • zepdrix
What do you get for the first set of blue brackets?
anonymous
  • anonymous
-8
anonymous
  • anonymous
n the second is the -8x e ^-4x^2
zepdrix
  • zepdrix
\[\huge y''=\color{orangered}{\left(-8\right)}e^{-4x^2}-8x\color{orangered}{\left(-8xe^{-4x^2}\right)}\] Like that? Yay good job!
anonymous
  • anonymous
i love that lol

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