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Chain Rule/Product Rule Question

  1. Aug 17, 2006 #1
    I am suppose to derive the following and then find x if f´(x)= 0

    f(x) = (6-x2) [tex]\bullet[/tex] e2x

    Which one should be used? I guess I am confused with the fact that it is e2x

    Product rule gives:

    (6-x2) [tex]\bullet[/tex] e2x + e2x [tex]\bullet[/tex] (-2x)

    e2x(6-x2-2x) (x1 = 0)

    -x2-2x+6 = 0 gives decimal values that doesn't seem to fit.

    Chain rule gives:

    I'm not even sure how to start here. I assume that the outer derivate is e2x? Or should the chain rule be applied to it and then the product rule?

    Some kind of hint as to what is the correct path would be greatly appreciated :smile:

    Edit: Latex mishap >_>
     
  2. jcsd
  3. Aug 17, 2006 #2

    VietDao29

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    Homework Helper

    I assume that the big, big dot in your post is the multiple sign.
    You can use both rules (i.e, Chain Rule, and Product Rule) in this problem. There's no limit of the number of the rules you can use. So, just use it where you think is appropriated.
    f(x) = (6 - x2) e2x
    f'(x) = (6 - x2)' e2x + (6 - x2) (e2x)' = ...
    You have taken the derivative of e2x with respect to x incorrectly.
    We should use the Chain Rule there. By letting u = 2x, we have:
    [tex]\frac{d}{dx} e ^ {2x} = \frac{d (e ^ u)}{du} \times \frac{du}{dx} = \frac{d(e ^ u)}{du} \times \frac{d(2x)}{dx} = ...[/tex]
    So, can you see where your mistake is?
    Can you go from here? :)
     
  4. Aug 17, 2006 #3
    Of course.

    The inner is 2.

    2e2x(6-x2)+ e2x(-2x)

    e2x(12-2x2)+ e2x(-2x)

    12-2x2-2x = 0

    and there we have the values of x.

    Thank you VietDao29, you were very helpful :smile:
     
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