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Q. f(x)=ln (12x-5/9x-2)So by using the chain rule, i can

  1. Sep 9, 2008 #1
    Q. f(x)=ln (12x-5/9x-2)

    So by using the chain rule, i can get:

    (-4/3)((9x-2)2/(12x-5)2)

    and by using the quotient rule, i can get the final answer, which is:

    (2(-36x-8)(-36x-15)2-2(-36x-15)(-36x-8)2)
    ------------------------------------------------------------------
    (-36x-15)4

    The answer that i got here is wrong, but i don't know, why?
     
  2. jcsd
  3. Sep 9, 2008 #2
    Re: Differentiation4

    First of all do you mean: [tex] f(x) = ln \left( \frac{12x-5}{9x-2} \right) [/tex]? In which case yes you would use chain rule and quotient rule. Chain rule applies to cases like f(g(x)), what is your f? what is your g? let's start there.
     
  4. Sep 9, 2008 #3
    Re: Differentiation4

    yes, and f is ln (12x-5/9x-2), where g is (12x-5/9x-2), correct??
     
  5. Sep 9, 2008 #4
    Re: Differentiation4

    Perhaps I used too many f's, let's say we have

    [tex] f(x) = g(h(x)) = ln \left( \frac{12x-5}{9x-2} \right) [/tex]

    What is g? what is h?
     
  6. Sep 10, 2008 #5
    Re: Differentiation4

    Thanks, but i solve the problem, the f(x) will be the nominator of the log and the g(x) will be the denominator of the log.
     
  7. Sep 10, 2008 #6
    Re: Differentiation4

    No, I asked about the chain rule, not the quotient rule. Actually f is the whole function, h(x) would be the argument of ln( ) and g would be ln itself. In general you should know this rule:

    If

    [tex] f(x) = ln( g(x) ) [/tex]

    Then

    [tex] f'(x) = \frac{g'(x)}{g(x)}[/tex]

    So in this case the derivative of

    [tex] ln \left( \frac{12x-5}{9x-2} \right) [/tex]

    is

    [tex] \frac{ \left(\frac{12x-5}{9x-2} \right)'} {\left( \frac{12x-5}{9x-2} \right) } [/tex]

    Where ' denotes differentiation.
     
  8. Sep 11, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Differentiation4

    Since this doesn't really have anything to do with differential equations, I am moving it to Calculus and Analysis.
     
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