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Finding f'(x)

  1. Jun 13, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x) = (x^2-3)/(7ln[x]+3) find f'(x)



    2. Relevant equations



    3. The attempt at a solution

    My attempt :
    (7ln[x]+3)(x^2-3)' - (x^2-3)((7ln[x]+3)' / (7ln[x]+3)^2

    (7ln[x]+3)(2x) - (x^2-3)((7/x) / (7ln[x]+3)^2

    (14xln[x]+6x) - ((7x^2-21)/x) / (7ln[x]+3)^2


    Is this right??

    I am so lost...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 13, 2010 #2

    Mentallic

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    Yeah it's right. Why do you feel so lost? :tongue:
     
  4. Jun 13, 2010 #3
    Thanks for the quick reply. Trying to study for finals but this did not feel like an adequate answer.
    Glad to hear I am on the right track.
     
  5. Jun 13, 2010 #4
    Do you have time for one more?
     
  6. Jun 13, 2010 #5
    f(x) = ln(5x^3 - 5)^3 find f'(x)

    My solution :
    u = (5x^3 - 5) so du is 15x^2
    ln(u)^3(du)

    3ln(5x^3 - 5)^2(15x^2) ??

    Is that right??
     
  7. Jun 13, 2010 #6

    Mentallic

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    No worries. Good luck on the exams!
     
  8. Jun 13, 2010 #7

    Mentallic

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    Oh that depends, is it [tex]ln\left((5x^3-5)^3\right)[/tex] or [tex]\left(ln(5x^3-5)\right)^3[/tex].

    From what you've done, it looks like the latter.
     
  9. Jun 13, 2010 #8
    it is y = ln(5x^3-5)^3
    so then the answer I had is right?
     
  10. Jun 13, 2010 #9

    Gib Z

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    That's not right. The derivative of ln(u)^3 is not 3ln(u)^2. Also, could you clarify whether you mean [tex]f(x) = \ln ( (5x^3-5)^3)[/tex] or [tex]f(x) = ( ln (5x^3-5) )^3[/tex].
     
  11. Jun 13, 2010 #10
    It is the second one you have posted... f(x) = (ln(5x^3-5))^3
    (how do you get the superscript?)
     
  12. Jun 13, 2010 #11

    Gib Z

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    Do you know how to simplify [tex] ln (a^b)[/tex]? That will help you for this question.
     
  13. Jun 13, 2010 #12

    Mentallic

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    Gib Z, the question is exponentiating the entire logarithm.

    For [tex]y=\left(ln(5x^3-5)\right)^3[/tex]

    Let [tex]u=5x^3-5[/tex]

    Now we have [tex]y=\left(ln(u)\right)^3[/tex]

    Let [tex]t=ln(u)[/tex]

    So finally we have [tex]y=t^3[/tex]

    Now the derivative is [tex]\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{du}.\frac{du}{dx}[/tex]
     
  14. Jun 13, 2010 #13
    Would it be bln(a) ?
     
  15. Jun 13, 2010 #14

    Mentallic

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    Yes ln(ab)=b.ln(a) but that's not the problem at hand, according to:

     
  16. Jun 13, 2010 #15
    Ok, I will use your previous post to figure it out. Thanks Mentallic! I really appreciate it!
     
  17. Jun 13, 2010 #16

    Mentallic

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    No problem :smile:

    By the way, I never use this long tedious method of substitution to find the derivative. Finding the pattern in what happens will really save you time and energy.

    Let's look at your most recent example:

    [tex]y=\left(ln(5x^3-5)\right)^3[/tex]

    First we picture the entire problem as (something)3, and so we use the usual rule and arrive at:

    [tex]\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2[/tex] --- not the answer

    but now we need to multiply everything by the derivative of what's inside, and then what's inside that... etc. till we reach a finishing point.

    we look at what's inside as being ln(something), so we have to multiply by the derivative of that, which, by our rule with logarithms becomes 1/(something).

    Now we have:

    [tex]\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2.\frac{1}{5x^3-5}[/tex] --- not the answer

    But again, we've ignored what's inside of that log, so we need to multiply by the derivative of that as well,

    So finally we end up with:

    [tex]\frac{dy}{dx}=3\left(ln(5x^3-5)\right)^2.\frac{1}{5x^3-5}.(15x^2)[/tex] --- the answer

    We can even go again and think of it as 5(something)3-5 but then the derivative of that something is 1 so it doesn't change anything. This is our true finishing point since there are no more variables to deal with.
     
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