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Homework Help: Derivative with general exponential and logarithmic fuctions

  1. Jun 26, 2008 #1
    1. The problem statement, all variables and given/known data

    f(x)=(((x^7)+3)((x^3)+2)) / (((x+3)^2)((x^6)+5))
    Find f'(x)

    2. Relevant equations

    d/dx lnx = 1/x
    d/dx ln(f(x)) = f'(x)/f(x)

    3. The attempt at a solution

    ln(f(x)) = ln((x^7)+3) + ln((x^3)+2) - ln((x+3)^2) - ln ((x^6)+5)
    f'(x)/f(x) = (7x^6)/((x^7)+3) + (3x^2)/((x^3)+2) - (2)/(x+3) - (6x^5)/((x^6)+5)

    f'(x) = [(((x^7)+3)((x^3)+2)) / (((x+3)^2)((x^6)+5))]*[(7x^6)/((x^7)+3) + (3x^2)/((x^3)+2) - (2)/(x+3) - (6x^5)/((x^6)+5)]

    My answer may be correct - just wanting someone to give it a quick check.

    I know it looks terrible just to even look at. If someone could please show me where/how to use better syntax I would greatly appreciate it. I'll even rewrite the problem so that it'll look more friendly :P
  2. jcsd
  3. Jun 28, 2008 #2


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    Science Advisor
    Homework Helper

    Hmm I missed out on this thread before, but actually that's a nice trick that I hadn't heard / thought of before.
    Actually this makes it very easy to prove that in general, the derivative of
    [tex]\phi(x) = \frac{f g}{h k}[/tex]
    (with f, g, h, k functions of x) is
    [tex]\phi' = \frac{f' g}{h k} + \frac{f g'}{h k} - \frac{f g}{h^2 k} h' - \frac{f g}{h k^2} k'[/tex]

    (And yes, your answer is correct)
  4. Jun 28, 2008 #3


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    Science Advisor

    It's just a fairly standard application of the logarithmic derivative. And yes, the answer is correct.
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