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Simple differentiation question

  1. Oct 19, 2003 #1
    "Simple" differentiation question

    There is a question which is as follows:

    (4^x)/(ln4)

    I have tried many times to get the answer but with no success, would someone be so kind as to show me how to do it?

    The answer is:

    4^x

    My working is as follows:

    f(x) = (4^x)/(ln4)
    F'(x) = ((ln 4 * ln 4 * 4^x) - (4^x - (1/4))) / ((ln 4)^2)
    f'(x) = (4^x) - ((4^x * (1/4)) / ((ln 4)^2))

    ...help?
     
  2. jcsd
  3. Oct 19, 2003 #2
    You cannot use quotient rule here.

    Let y = 4x

    take ln on both sides

    ln y = x ln 4

    Take the first derivatives on both sides

    y'/y = ln 4

    y' = y ln 4 = 4x ln 4

    So,

    F(x) = (4^x)/(ln4)

    F'(x) = (1/ln 4) * y' (since the denominator is a constant) = 4^x
     
  4. Oct 19, 2003 #3

    Hurkyl

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    Well, you can use the quotient rule here if you remember that the derivative of a constant (like ln 4) is zero.

    (of course, you should just factor the constant out like you suggested)
     
  5. Oct 20, 2003 #4
    Ahh yip, got it. Thanks guys :D
     
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