evaluating 2nd derivative of y=2x*e^x when x =-1

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Dummy83

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Apr 18, 2017
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Good evening ladies and gentlemen

im having a lot of trouble understanding how to start or finish this question
Evaluate the function of the second derivative of y=2x*e^x where x = -1
any help would be greatly appreciated!
Thanks
Vince
 
Dummy83 said:
Good evening ladies and gentlemen

im having a lot of trouble understanding how to start or finish this question
Evaluate the function of the second derivative of y=2x*e^x where x = -1
any help would be greatly appreciated!
Thanks
Vince
Click to expand...
Start with deriving the second derivative of the given function. What do you get?
 
You need to start by taking the first derivative! You have \(\displaystyle f(x)= 2xe^{x}\). That is the product of \(\displaystyle g(x)= 2x^2\) and \(\displaystyle h(x)= e^x\) so use the "product rule": (gh)'= g'h+ gh'.

Once you have that, differentiate again.
 
In addition to the very cogent replies you have already received, I find that beginning students do not grasp that derivatives are functions in their own right. If I told you to find \(\displaystyle g(-\ 1) \text { given } g(x) = x^2\), you would have no trouble.

All this problem is doing is saying

\(\displaystyle g(x) = f''(x) \text { and } f(x) = 2x * e^x \implies g(-\ 1) = what?\)
 
JeffM said:
In addition to the very cogent replies you have already received, I find that beginning students do not grasp that derivatives are functions in their own right. If I told you to find \(\displaystyle g(-\ 1) \text { given } g(x) = x^2\), you would have no trouble.

All this problem is doing is saying

\(\displaystyle g(x) = f''(x) \text { and } f(x) = 2x * e^x \implies g(-\ 1) = what?\)
Click to expand...

I didn't get that...:(
 
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