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Math Help - applying chain rule to natural log

  1. #1
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    applying chain rule to natural log

    Hi, I'm very confused about these:
    1. y=2^ln x
    2. y=ln2^x

    for #1, wouldn't it be:
    y=2^ln x
    y'=ln x (2^ln x) (ln 1/x) ?

    but the answer in the book says it's y'=ln x (2^ln x) (1/x).

    and I have absolutely no idea how to begin #2.

    edit:
    sorry, but these too:
    3. f(x) = Ae^-Bx
    book says f'(x) = Ae^-Bx (-B). I don't understand where would the -Bx go
    4. f(x) = L/(1+Ae^-Bx)

    thank you so much in advance!
    Last edited by colloquial; October 7th 2009 at 02:10 PM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by colloquial View Post
    Hi, I'm very confused about these:
    1. y=2^ln x
    2. y=ln2^x

    for #1, wouldn't it be:
    y=2^ln x
    y'=ln x (2^ln x) (ln 1/x) ?

    but the answer in the book says it's y'=ln x (2^ln x) (1/x).

    and I have absolutely no idea how to begin #2.

    thank you so much in advance!
    For number 1 use the property that \frac{d}{du}a^u = a^u\,ln(u)

    a = 2, u = ln(x)

    \frac{d}{dx}(2^{ln(x)}) = 2^{ln(x)}\,ln(ln(x)).

    Perhaps 1/ln(x) is obtained from the chain rule but I don't see it.


    For number 2 use the property of logs that says ln(b^a) = a\,ln(b)

    y = ln(2^x) = x\,ln(2)

    This should be easy to differentiate. Remember ln(2) is a constant
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    For number 1 use the property that \frac{d}{du}a^u = a^u\,ln(u)
    I would suggest you NOT use that property as it isn't true! The correct property is \frac{a^u}{du}= ln(a) a^u. Write a^u= e^{ln(a^u)}= e^{uln(a)} and use the chain rule: the derivative is e^{u ln(a)} times the derivative of u ln(a) with respect to u which is ln(a), not ln(u).

    a = 2, u = ln(x)

    \frac{d}{dx}(2^{ln(x)}) = 2^{ln(x)}\,ln(ln(x)).

    Perhaps 1/ln(x) is obtained from the chain rule but I don't see it.
    The derivative of 2^{ln(x)} is, by the chain rule, again, ln(2) 2^{ln(x)} times the derivative of ln(x) which is 1/x. The derivative of 2^{ln(x)} is ln(2)\frac{2^{ln(x)}}{x} so your book (believe it or not!) is correct.

    For number 2 use the property of logs that says ln(b^a) = a\,ln(b)

    y = ln(2^x) = x\,ln(2)

    This should be easy to differentiate. Remember ln(2) is a constant
    Exactly right!
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