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Derivative of natural log function questions

  1. Mar 30, 2015 #1
    1. The problem statement, all variables and given/known data


    Hello, I had a few derivative of the natural logarithm functions questions. It seems like it should be fairly straightforward, but I am turning it into a pig’s ear.


    On my honor, none of these are problems on an assessment per se, however, they are not materially different than anything that I am finding in my calc book and what I would assume to be on a midterm or final:

    2. Relevant equations

    a) ln 3x

    b) ln 3x + x

    c) ln (3x + 4)

    d) ln (x^2)

    e) ln x^2

    f) x (ln 2x)^4

    3. The attempt at a solution

    a) I believe I got this right – 1/3x

    b) I think I have this one as well – 1/3x + 1

    c) I have (3x + 4)’/(3x + 4) = 3/(3x + 4)

    d) Using the chain rule, I believe this is 2x/x^2

    e) I believe using the ln x derivative rules, this is 2 (ln x)’ = 2/x

    f) This is giving me problems. Using the product rule and chain rule, I think the answer

    = 1 * (ln 2x)^4 + x * 4(ln 2x)^3 * 1/2x

    = (ln2x)^4 + 4 [(ln 2x)^3]/x, with an x in the numerator and denominator in the final term cancelling each other out.

    Thank you for your help --

    SY
     
  2. jcsd
  3. Mar 30, 2015 #2

    Mark44

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    Maybe, but you didn't write it correctly. I think you mean 1/(3x) but what you wrote is (1/3)x.

    Also, you need to use the chain rule.
    As above. This should be written as 1/(3x) + 1. You need to use the chain rule, though.
    Looks good
    This should be simplified. Since ln(x2 = 2ln(x), its derivative is 2/x.
    Aren't d and e the same question?
    That is, unless you mean (ln(x))2
    Did you forget to write the factor of x in the numerator?
    I agree that the x factors in the numerator and denominator cancel.
     
    Last edited: Mar 30, 2015
  4. Mar 30, 2015 #3

    LCKurtz

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    Is that what you meant or do you mean ln(3x+x)? Just curious, could be either way.

    Nope. Needs chain rule, even if you meant 1/(3x).
    Nope Same problem.
     
  5. Mar 30, 2015 #4
    a) I forgot to write it as 1/ (3x) -- it looked better handwritten on paper.

    b) Same traffic -- it should say [1/(3x)] + 1

    c) Woo-hoo! -- basically a thru c are good to go.

    d) I have that written as ln (x^2) -- that is where I had the chain rule as giving a derivative of 2x/(x^2) = 2/x

    e) That one should be (ln x)^2 -- 2 ln x, again with an answer of 2/x

    f) y = x(ln 2x)^4

    Not sure if I missed an x in the first attempt, but here, I still have the product rule and chain rule as:

    y' = 1 * (ln 2x)^4 + x * 4(ln 2x) ^3 * 1/(2x)

    y' = (ln 2x) ^4 + 4x (ln 2x) ^3 / (2x) -- the 4x in the numerator and 2x in the denominator cancel out, with a numerator of 2 left

    y' = (ln 2x)^4 + 2 (ln 2x)^3

    Thoughts?

    Thanks for all help

    SY
     
  6. Mar 30, 2015 #5

    Dick

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    Let's just start with the first one. As LCKurtz has already pointed out, the derivative is not 1/(3x). Fix that one first.
     
  7. Mar 30, 2015 #6

    SammyS

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    f) now looks good.

    - but (e) is another matter.
    ## \left(\ln(x)\right)^2\ne 2\ln(x)##

    Use the chain rule. This is much like the ##\ \left(\ln(2x)\right)^4\ ## portion of part (f), but simpler.
     
  8. Mar 30, 2015 #7
    y = (ln x)^2

    y' = 2 ln x * 1/x

    y' = (2 ln x)/x

    I think I got it this time.

    Also, another above problem --

    y = ln 3x

    y' = 1/(3x) -- correct or not?

    SY
     
  9. Mar 30, 2015 #8

    Dick

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    First one correct, second one not correct. Try the chain rule.
     
  10. Mar 30, 2015 #9
    y' = [1/(3x)] * 3 = 3/(3x) = 1/x
     
  11. Mar 30, 2015 #10

    Dick

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    Now it's right. You could also write ln(3x)=ln(3)+ln(x). Now it's pretty obvious the derivative is just 1/x.
     
  12. Mar 30, 2015 #11
    Thanks to all for the help. Derivatives of ln x and I had a long day today.
     
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