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Math Help - Help with derivative of y= ln x

  1. #1
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    Question Help with derivative of y= ln x

    So I have this problem solved algebraically, but need to use derivatives...
    Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
    The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

    ln(ab) = -ln(1/ab)
    = -ln(1/a * 1/b)
    = -[ln(1/a) + ln(1/b)]
    = -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
    this is what I have come up with, but it is not right...
    I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
    Any help out there would be greatly appreciated. Thanks for looking...
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  2. #2
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    Quote Originally Posted by bemidjibasser View Post
    So I have this problem solved algebraically, but need to use derivatives...
    Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
    The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

    ln(ab) = -ln(1/ab)
    = -ln(1/a * 1/b)
    = -[ln(1/a) + ln(1/b)]
    = -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
    this is what I have come up with, but it is not right...
    I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
    Any help out there would be greatly appreciated. Thanks for looking...
    Consider the functions

    y_1=\ln(a)+\ln(x) \implies y_1'=\frac{1}{x}

    y_2=\ln(ax) \implies y_2'=\frac{1}{ax}(a)=\frac{1}{x}

    So the two functions can differ by at most a constant C

    y_1=y_2+C

     \ln(a) +\ln(x)=ln(ax)+C this is true for all x so let x=1

     \ln(a) +\ln(1)=ln(a)+C \iff C=0

    So sub back in to get

    \ln(a)+\ln(x)=\ln(ax) for all x.

    Yay
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by bemidjibasser View Post
    So I have this problem solved algebraically, but need to use derivatives...
    Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).
    The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

    ln(ab) = -ln(1/ab)
    = -ln(1/a * 1/b)
    = -[ln(1/a) + ln(1/b)]
    = -ln(1/a) + -ln(1/b) = ln(a) + ln(b)
    this is what I have come up with, but it is not right...
    I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
    Any help out there would be greatly appreciated. Thanks for looking...
    Using the chain rule and remembering that b is a constant, we have:

    \frac{d}{da}[\ln(ab)] = \frac{1}{ab}\cdot\frac{d}{da}[ab] = \frac{b}{ab} = \frac{1}{a}

    \frac{d}{da}[\ln(a)+\ln(b)]=\frac{1}{a}

    So they have the same derivative. Use b=1 and solve \ln(ab)=\ln(a)+\ln(b)+c.
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