1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Staff Emeritus
    Science Advisor

    It's just a fairly standard application of the logarithmic derivative. And yes, the answer is correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?