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D/dx 4ln3x=4x - how?

  1. Oct 5, 2012 #1
    Would like help with this question

    Question:y = 4ln3x, find dy/dx


    2. Relevant equations

    1. I used the chain rule:

    y=4ln u and u =3x

    dy/du=4/u THIS IS MY PROBLEM

    There is no number between the ln and u, eg. ln4u, so WHY do you use
    d/dx ln(ax+b)=a/(ax+b)?

    3. The attempt at a solution

    y=4ln u and u =3x

    dy/du=4/u du/dx=3

    dy/dx=12/3x=4x.
     
  2. jcsd
  3. Oct 5, 2012 #2
    If I got it right, you mean [itex]\displaystyle \frac{d}{dx}4\log(3x) = 4\frac{d}{dx}\log(3x)[/itex].

    Now, it should be already plain obvious to some people what the answer to this is, but in case it isn't, there are two ways to go here. One is to use the fundamental identity of logarithms (how does a logarithm distribute over a product?), and the other is to use the chain rule. Pick whichever you want.
     
  4. Oct 5, 2012 #3

    jedishrfu

    Staff: Mentor

    your last eqn: dy/dx=12/3x=4x should be = 4/x right?
     
  5. Oct 5, 2012 #4

    Mark44

    Staff: Mentor

    dy/du = 4/u
    dy/dx = dy/du * du/dx = 4/u * 3 = ?
     
  6. Oct 5, 2012 #5

    SammyS

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    You might also notice that [itex]4\ln(3x)=4(\ln(3)+\ln(x))[/itex]
    [itex]=4\ln(3)+4\ln(x)\ .[/itex]​
     
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