1. Limited time only! Sign up for a free 30min personal 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!

Logarithm Derivatives

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Find f'(x): [itex] (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3)) [/itex] <- Not sure why it''s not showing up but the 1/3 is an exponent to just the (3x+2).


    2. Relevant equations
    Product Rule and Quotient Rule for Differentiating


    3. The attempt at a solution
    So I thought I should split it into two parts: A and B.

    A) [itex] (x^2)/((2x-3)^2) [/itex]
    B) [itex] ((3x+2)^1/3)/(2x-3) [/itex]

    My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives me the original denominator power (4).

    I end up getting for A) (-12x^2 + 18x)/((2x-3)^4)

    For B, I have something which seems much more complex, because of the negative fractional exponents, and this is where I'm stuck. I want to simplify, but I'm not sure how to without screwing everything up. This is my last step for B:

    [itex] ((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^2) [/itex] <- The -2/3 is the exponent

    I really appreciate the help. And also, as for the absolute value part of the question, well I haven't even thought about that- I'm stuck on the easy part itself. -_-

    Thank you!
     
    Last edited: May 24, 2013
  2. jcsd
  3. May 24, 2013 #2

    Mark44

    Staff: Mentor

    In LaTeX, if an exponent is more than one character, you have to put braces around the entire expression of the exponent. If your exponent expression includes parentheses, they need to go inside the braces as well.

    So (3x + 2)1/3 would be written as (3x + 2)^{1/3}.
    No, you can't do this. It's not true that d/dx( f(x) * g(x)) = f'(x) * g'(x). That seems to be what you're trying to do, from what you're saying below.
    d/dx |x| = x/|x|, and
    d/dx |u| = u/|u| * du/dx
     
  4. May 24, 2013 #3
    Oh wow, that was the rule that we learned in school- I should have known I'm not supposed to do that! I'll redo the problem, and see what I get. Thanks!
     
  5. May 24, 2013 #4
    If [itex] d/dx [x] = x/|x| [/itex] ,

    then would the correct expression be [itex] d/dx |x| = (f(x))/|((x^2)((3x+2)^{1/3}))/((2x-3)^3)| [/itex] ?

    meaning, in simplified form would it be: [itex] ((f(x)) * ((2x-3)^3)/((x^2)(3x+2)^{1/3}) [/itex]

    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Logarithm Derivatives
  1. Logarithmic derivative (Replies: 3)

  2. Logarithm derivative (Replies: 4)

Loading...