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Regarding logarithmic differentiation

  1. Nov 3, 2013 #1
    Thank you for viewing my thread. I have been given the following steps for logarithmic differentiation:
    1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.
    2. Differentiate implicitly with respect to x.
    3. Solve the resulting equation for y'.

    I was wondering if I could go about this in another way.
    1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.
    2. At this point I would have something like:
    ln(y) = a+b+c
    Instead of doing implicit differentiation , could I do this:
    e^[ln(y)] = e^(a+b+c)
    y = e^(a+b+c)

    Thanks.
     
    Last edited: Nov 4, 2013
  2. jcsd
  3. Nov 3, 2013 #2

    fzero

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    The problem with this is that ##e^{a+b+c}## is just equal to ##f(x)##. The laws of logarithms that you use to simplify ##\ln f(x)## don't make ##f(x)## any simpler than using the rules for simplifying exponents and products. The reason to take the log is to make differentiating things like ##(x+2)^x## a bit simpler, since it is easier to differentiate ##x \ln (x+2)##.
     
  4. Nov 4, 2013 #3
    Yeah I get what you're saying now. I completely forgot a step lol. For some reason the first time I looked at it I forgot that I was doing differentiation. In my mind I had it as:
    y' = e^(a+b+c)

    Thank you!
     
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