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Dereiv. of 4^x + 3^x + 9^-x

  1. Nov 9, 2004 #1
    if anyone can teach me how to do this that would be great, thanks.

    dereiv. means derivative sorry
     
  2. jcsd
  3. Nov 9, 2004 #2
    when im doing something with the form something raised to x i always remember - keep the tern , log (or ln, same meaning here) the base number

    and then differentiate the exponent

    for example for [tex] \frac{d}{dx} (3^x) = 3^x Log3 (1) [/tex]

    as you can see keep the function 3^x, log the base Log3, and then differentiate teh numerator (1).
     
  4. Nov 9, 2004 #3

    Zurtex

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    All I can say to the above post, is eh? That would mean that it would be 0. In general:

    [tex]\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x[/tex]

    Where a is some constant. Here is the method used to work it out and generally useful for this type of problem:

    [tex]y= a^x[/tex]

    [tex] \ln y = \ln \left( a^x \right)[/tex]

    [tex]\ln y = x \ln a[/tex]

    [tex]\frac{dy}{dx} \frac{1}{y} = \ln a[/tex]

    [tex]\frac{dy}{dx} = (\ln a)y[/tex]

    [tex]\frac{d}{dx} \left( a^x \right) = \ln (a) \; a^x[/tex]
     
  5. Nov 9, 2004 #4
    what would be zero??
     
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