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Derivative of log

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    find f '(x) of 5^xsinhx


    2. Relevant equations
    i feel like im missing somthing can someone direct me to the right direction pleasee???


    3. The attempt at a solution
    product rule (f ')(g)+(f)(g ') f=5^x g=sinhx

    5^xsinhx= e^ln(5)^xsinhx

    (e^ln(5)^x)(sinhx)+(5^x)(coshx)(ln5)

    (e^ln(5)^x)(sinhx)+(5^x)(ln5)(coshx)??????
     
  2. jcsd
  3. Feb 13, 2009 #2

    nicksauce

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    Well it looks like you took the derivative of none of the terms in the first term, and both of the terms in the second term. Try to fix that.
     
  4. Feb 13, 2009 #3
    The factor of ln(5) is in the wrong term.

    Write them out separately:
    f =
    f' =
    g =
    g' =

    Then put them together.

    By the way, ensure you use parentheses to remove ambiguity when typing exponents; e.g., 5^x = (e^(ln(5)))^x = [tex]\left(e^{\ln{5}}\right)^x[/tex] etc.
     
  5. Feb 13, 2009 #4
    "none of the terms in the first term" what do you mean by this?
     
  6. Feb 14, 2009 #5

    Mark44

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    As Unco pointed out, your expression above is ambiguous. It's not clear what the exponent on 5 is. It could be x, or it could be x*sinh(x). In other words, is this function (5^x)* sinh(x) or is it (5^x)^sinh(x)?
    Now from this work it appears that the function is (5^x) * sinh(x), in other words, as the product of two functions. Inasmuch as you're using the product rule, you must think that the function is (5^x)*sinh(x).
    Now it looks like an exponential function raised to a power, so the function is not a product, which means that the product rule does not apply.

    What exactly is the function? We can't help you if you don't know what it is.
     
  7. Feb 14, 2009 #6

    HallsofIvy

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    5^x is NOT "e^ln(5)^x" it is e^(x ln(5))

    Which means that (5^x)'= (e^(x ln(5)))'= ln(5) e^(x ln(5))= ln(5) 5^x.
     
  8. Feb 14, 2009 #7
    here is the original problem find f '(x).

    f(x)=5^xSinhx does this clarify???
     
  9. Feb 14, 2009 #8

    tiny-tim

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    oooh, why can't everyone learn how to use the X2 tag (just above the Reply box)? :cry:

    It's e(ln5)xsinhx.​
     
  10. Feb 16, 2009 #9
    yea and thats what i put on top??? so since the product rule is (f ')(g)+(f)(g ')
    so wouldnt the answer be (e^ln(5)xsinhx+(5^x)(coshx)?????
     
  11. Feb 16, 2009 #10

    tiny-tim

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    nooo! try the X2 tag! :rolleyes:
    product rule?

    what product? :confused:

    use the chain rule first! :smile:
     
  12. Feb 16, 2009 #11
    i don't think its the right answer e^sinhx(coshx)
     
    Last edited: Feb 16, 2009
  13. Feb 17, 2009 #12
    ok so i got 5^xcoshx+5^x(ln5)sinhx?????????????????
     
  14. Feb 17, 2009 #13

    tiny-tim

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    not even close :cry:

    show us your full working …

    differentiate e(ln5)xsinhx using the chain rule :smile:
     
  15. Feb 17, 2009 #14
    i used the chain rule and i got e^(ln5xsinhx)(ln5)???????
     
  16. Feb 17, 2009 #15

    tiny-tim

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    what is the derivative of (ln5)xsinhx?
     
  17. Feb 17, 2009 #16
    1/5xcoshx
     
  18. Feb 17, 2009 #17

    Mark44

    Staff: Mentor

    It looks like you are doing this: d/dx(ln 5) = 1/5

    If so, that's wrong.

    Also, when you differentiate (ln 5)* x*sinh(x), you should be getting two terms.

    The answer in your previous post might be interpreted in several ways:
    [tex]\frac{1}{5x cosh(x)}[/tex]
    [tex]\frac{1}{5x}cosh(x)[/tex]
    [tex]\frac{1}{5}xcosh(x)[/tex]

    Use parentheses or learn LaTex!
     
  19. Feb 17, 2009 #18
    ln5*sinh(x)+ln5*x*cosh(x)?????????
     
  20. Feb 17, 2009 #19

    Mark44

    Staff: Mentor

    Yes, that's it. Now you can respond to Tiny Tim's request in post 13.
     
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