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Math Help - Derivatives - product rule with three terms?

  1. #1
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    Derivatives - product rule with three terms?

    I'm trying to get the derivative of 3xe^(2x)

    I know the product rule and I know the derivative of e^2x is 2e^(2x).
    So I was going to use the product rule, but I've only ever done that with two terms. Here, there are three.
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  2. #2
    Senior Member Pinkk's Avatar
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    You have a function h(x)=f(x)g(x) where f(x)=3x and g(x)=e^{2x}

    h'(x)=f'(x)g(x)+f(x)g'(x)

    h'(x)=(3)e^{2x}+3x(2e^{2x})

    h'(x)=3e^{2x}+6xe^{2x}
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  3. #3
    MHF Contributor matheagle's Avatar
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    the 3 is just a constant, BUT if you have three functions then

    (fgh)'=f'gh+fg'h+fgh'.
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  4. #4
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    thats a product rule and a chain rule

    3xe^(2x)

    3e^(2x)

    for the first. take derivative of 3x which is just 3 and keep second part. than. chain rule e^(2x) so e^ whatever is just e^ the whatever. so e^(2x) than multiply by 2 since the derivative of 2x is 2. keep the first part.

    So you get:
    3e^(2x) + 3xe^(2x)*2 which is 3e^(2x) + 6xe^(2x)
    you can then simplify by pulling 3e^(2x) out of the answer
    3e^(2x)*(1+2x) but that wouldnt really help unless your graphing and looking for critical numbers
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  5. #5
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    Just in case a picture helps...



    You have the chain rule - generally this pattern -



    ... wrapped inside the product rule...



    ... more visible here if we enclose some of the balloons in some more...



    Hope this helps, or doesn't further confuse.
    Don't integrate - balloontegrate!
    Balloon Calculus: worked examples from past papers
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