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Derivative of e^2

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Derivative of f(x) = x3 + e2


    2. Relevant equations

    Dex = ex

    D constant = 0

    3. The attempt at a solution

    f'(x) = 3x2 + 0?

    Is e2 treated as a constant?
     
  2. jcsd
  3. Mar 23, 2010 #2
    Yes. Or you can use the chain rule. if u = f(x) = 2 and y = g(u) = [itex] e^u [/itex] then

    [tex] \frac {dy} {dx} = \frac {dy} {du} \frac {du} {dx} [/tex]

    since [tex] \frac {du} {dx} = 0 [/tex] [tex] \frac {dy} {dx} = 0 [/tex].
     
  4. Mar 23, 2010 #3

    Mark44

    Staff: Mentor

    Not only is it treated as a constant, it is a constant. The derivative of any constant is zero. Period.

    Using the chain rule certainly works, but it's definitely overkill, so not recommended.
     
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