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B Derivation with logarithms and product

  1. May 27, 2017 #1
    I went through an example question that showed me how to solve the question but I'm not sure if I've misunderstood something or if they did a mistake.

    Question: Derivate y = (1/ax)ax

    ln(y) = ln( (1/ax)ax ) = ax( ln(1) - ln(ax) ) = -ax ln(ax)
    (1/y)(dy/dx) = -ax * ax ln(a) - a * ln(ax)
    dy/dx = (1/ax)ax * (-ax * ax ln(a) - a * ln(ax))

    I think that they are derivating -ax ln(ax) as a product of two functions with u*(dv/dx) + v*(du/dx) where u = -ax and v = ln(ax). But isn't (dv/dx) supposed to be only ln(a)? So dy/dx = (1/ax)ax * (-ax * ln(a) - a * ln(ax))
  2. jcsd
  3. May 27, 2017 #2


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    Your derivation simplifies a lot if you use the fact that [itex]ln(a^x) = x ln(a)[/itex]
  4. May 27, 2017 #3


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    Try to differentiate ##\ln a^x## step by step.
    What is ##\frac{d}{dx}\ln f(x)## and how do you differentiate ##f(x)=a^x##?

    Hint: You can write ##f(x)=a^x=\exp(\ln a^x)=\exp(x\ln a)##.
  5. May 28, 2017 #4
    But isn't v = ln(ax) = x ln(a)
    dv/dx = ln(a)

    I can get that dv/dx is ax ln(a) if v = ax so:
    v = ax -> ln(v) = ln(ax) = x ln(a)
    (1/v)(dv/dx) = ln(a) -> dv/dx = v ln(a) = ax ln(a)

    But v = ln(ax) so I can't get anything but ln(a)
  6. May 28, 2017 #5


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    You're right. I first mistakenly forgot a denominator. Maybe the author you have your example from made the same mistake.
  7. May 28, 2017 #6


    Staff: Mentor

    Although "derivate" is a word in English, it is not used in mathematics. To obtain the derivative of a function, you differentiate it or find its derivative.
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