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Derivatives of Exponential Functions Question

  1. Nov 13, 2006 #1
    The problem is:

    y= 6^3x-7
    Read: Y equals 6 to the 3x minus 7 OVER/Divided by x to the 2nd minus x

    Now I thought I had to do a quotient rule there and start by doing:

    (x^2-x)(e^(ln6)(3x-7))(3) and then so on from there. I just don't know what to do with that 6 to the 3x-7. Is it something to do with "e"?
  2. jcsd
  3. Nov 13, 2006 #2


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    [tex]6^{3x-7} = e^{(3x-7)*ln(6)}[/tex] You can see this by rewriting it as [tex](e^{ln(6)})^{3x-7}[/tex] although the first form is easier to differentiate. So the derivative, by the chain rule, is [tex]e^{3x-7)*ln(6)}*ln(6)*3[/tex] or [tex]6^{3x-7}*3ln(6)[/tex]
  4. Nov 13, 2006 #3
    So would the quotient rule look like:

    (x^2-x)(e^(3x-7)(ln 6))(3ln 6) + (6^3x-7)(2x-1)

    By the way, how do I get my question to look like nice and neat without a bunch of ^ like you just did?
  5. Nov 13, 2006 #4
    Is a mathematical markup language called LaTeX.

    Here's some more info for this forum's LaTeX engine:

  6. Nov 14, 2006 #5


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    More specifically, the derivative of ax is ln(a) ax.

    If you click on something like

    [tex]\frac{da^x}{dx}= ln(a) a^x[/tex]
    A popup box will show the code.
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