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Derivative of power function

  1. Oct 9, 2013 #1

    Qube

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    Derivative of exponential function

    1. The problem statement, all variables and given/known data

    f(x) = 4^x + e^(2tanx)

    find f'(0)

    2. Relevant equations

    I used log differentiation.

    y = 4^x + e^(2tanx)

    lny = xln4 + 2tanx(lne) = xln4 + 2tanx

    3. The attempt at a solution

    y'/y = ln4 + 2sec^2(x)

    Solving for y I get y = 2.

    Plugging in x = 0 I get:

    y'/2 = ln4 +2sec^(0) = ln4 + 2(1)

    y' = 2(ln4+2)

    This however, isn't the answer.
     
    Last edited: Oct 9, 2013
  2. jcsd
  3. Oct 9, 2013 #2

    Mark44

    Staff: Mentor

    No - don't do this yet. Solve for y' algebraically.

    There's a difference between y' and y'(0). The first is a function, and the second is a number. Your need the function.
     
  4. Oct 9, 2013 #3

    Mark44

    Staff: Mentor

    Btw, your title is misleading. 4x is NOT a power function. It's an exponential function.

    In a power function, the variable is in the base, and the exponent is constant. E.g., x3, y5, and so on are power functions.
     
  5. Oct 9, 2013 #4

    Qube

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    Alright.

    y' = y(ln4 + 2sec^2(x))

    How do I proceed? Do I plug in y =2? If so I still get the same answer if I plug in x = 0.
     
  6. Oct 9, 2013 #5

    LCKurtz

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    This step in incorrect. ##\ln(a+b) \ne \ln a + \ln b##.
     
  7. Oct 9, 2013 #6

    Qube

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    I got it now! Thanks! I should probably review my log rules. It's amazing that after two years of calculus and a 5 on both AP Calc AB and BC I still make the simplest of pre-calculus mistakes!
     
  8. Oct 9, 2013 #7

    Ray Vickson

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    Your second line essentially says that
    [tex]\log(a+b) = \log(a) + \log(b)\;\leftarrow \text{ false!}[/tex]
    What IS true is ## \log(a\cdot b) = \log(a) + \log(b)##.
     
  9. Oct 9, 2013 #8

    Mark44

    Staff: Mentor

    I glossed right over that in the OP's work. Oh, well.
     
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