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Math Help - Help on differentiation

  1. #1
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    Help on differentiation

    Hello to everyone, I have just signed up and hope that I can grow welcome on these forums.

    I have a calculus assignment due in this week, and needed some help on a few questions, any help would be much appreciated.

    Only need help on 2)e)


    ---------------------

    Help on this entire question is needed


    Thanks ,
    Matt.
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  2. #2
    Eater of Worlds
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    2e is just a matter of the product rule.

    f(x)=e^{x}

    g(x)=ln(2x)

    Now, implement the rule.

    f(x)g'(x)+g(x)f'(x)
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  3. #3
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    Though, to differentiate ln f(x) is it:

    f'(x)
    ----
    f(x)
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  4. #4
    Eater of Worlds
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    Just use the chain rule.

    \frac{1}{2x}\cdot{2}
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Phatmat View Post
    Help on this entire question is needed


    Thanks ,
    Matt.
    for 3(a), use the chain rule, the result follows immediately, you may want to write out some steps though, so it looks like you're doing something. remember... \frac d{dx} h(k(x)) = h'(k(x)) \cdot k'(x)

    that is, you differentiate the entire function leaving the inside function intact (that is, you treat it as if it were a single variable), then multiply by the derivative of the inside function

    it's awkward using h and k as the functions for this rule, but f was taken, so...

    for 3(b): the product rule says: \frac d{dx}h(x) \cdot k(x) = h'(x)k(x) + h(x)k'(x)

    just apply the product rule here (use part (a) to find the derivative of [g(x)]^{-1}). simplify, and you will get the quotient rule


    Quote Originally Posted by Phatmat View Post
    Though, to differentiate ln f(x) is it:

    f'(x)
    ----
    f(x)
    yes. as galactus said, we can see this by using the chain rule
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