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Math Help - chain rule & inverse functions...derivatives

  1. #1
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    chain rule & inverse functions...derivatives

    I did this problem, and came up with the wrong answer. Just needed help on what i did wrong...thanks

    f(x) = eln(e^2x2 + 3)

    i got 4x(e^x) and it was way wrong...some help
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathaction View Post
    I did this problem, and came up with the wrong answer. Just needed help on what i did wrong...thanks

    f(x) = eln(e^2x2 + 3)

    i got 4x(e^x) and it was way wrong...some help
    do you mean f(x) = e^{\ln \left( e^2 x^2 + 3\right)} or is it actually f(x) = e \ln \left( e^2 x^2 + 3\right) ?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    do you mean f(x) = e^{\ln \left( e^2 x^2 + 3\right)} or is it actually f(x) = e \ln \left( e^2 x^2 + 3\right) ?
    sorry, its

    e^(lne^(2x^2)+3)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathaction View Post
    sorry, its

    e^(lne^(2x^2)+3)
    ok, so the +3 in the power of the first e is not a part of the power of the second e, got it. here's what we do. we would simplify first:

    y = e^{\ln e^{2x^2} + 3}

    \Rightarrow y = e^{2x^2 + 3}

    \Rightarrow y' = 4xe^{2x^2 + 3}
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  5. #5
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    ok...i just have to remember that whenever i have an e^x that expression will be the same in the derivative as the original function
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