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Simple Question - Derivative of log(coshx-1)

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    f(x) = Log(cosh(x-1)), find f'(x).

    2. Relevant equations

    3. The attempt at a solution
    f'(x) = [1/cosh(x) - 1] * d/dx [cosh(x) - 1], => f'(x) = sinh(x) / [cosh(x) - 1]

    Although, my marking scheme says the answer should instead be; cosh(x) + 1 / sinh(x).

    Can someone explain where I'm going wrong? Thanks.
  2. jcsd
  3. May 3, 2017 #2


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    Check the function f(x) in the problem. Is it really ##f(x)=\log(\cosh(x-1))##?
  4. May 3, 2017 #3
    Screenshot (56).png

    Question 16. Definitely yeah.
  5. May 3, 2017 #4


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    log(cosh(x)-1) is not the same as log(cosh(x-1)). The argument of cosh is x in the original problem, you wrote(x-1).
    You made errors with the parentheses in the solution. Somehow you wrote the correct derivative, but the marking scheme is not correct.
  6. May 3, 2017 #5
    Ah apologies, I tend to overuse brackets & at times make mistakes as a result. I'm working from a University past paper so I assumed the marking scheme to be correct no matter what as it's a paper from 2014.

    So in either circumstances with the parenthesis, there is no way that function could have a derivative equal to the marking schemes answer?
  7. May 3, 2017 #6


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    The derivative of f(x)=log(cosh(x)-1) is ##\frac{sinh(x)}{\cosh(x)-1}## as you wrote, and that is not among the given answers.
  8. May 3, 2017 #7


    Staff: Mentor

    Overuse is not the problem here -- [1/cosh(x) - 1] -- in the righthand side of the first equation above.
    What you wrote means ##\frac 1 {\cosh(x)} - 1##, which I'm sure isn't what you intended.

    When you write a fraction where either the numerator or the denominator contains multiple terms, you have to surround that part with parentheses or other enclosing symbols.
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