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Math Help - Bit of help please, rule

  1. #1
    ADY
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    Bit of help please, rule

    Ok, so how to do we use the quotient rule to show that this function

    g(x) = \frac{e^{(20-x/36)}}{(5+x)^2} ................ (x>-5)

    has derive.

     k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}

    Thank you
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  2. #2
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    Quote Originally Posted by ADY View Post
    Ok, so how to do we use the quotient rule to show that this function

    g(x) = \frac{e^{(20-x/36)}}{(5+x)^2} ................ (x>-5)

    has derive.

     k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}

    Thank you
    Let f= e^{20-x/36} and let g=(5+x)^2 then use the quotient formula \frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2} and simplify.
    Keep in mind that the derivitive of (5+x)^2 is 2(x+5) and f'(x)= \frac{1}{36}e^{20-x/36} (you can verify). So you have:

    \frac{[(5+x)^2 . \frac{e^{20-x/36}}{36}] - [e^{20-x/36} . 2(x+5)]}{(5+x)^4} = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}
    Last edited by Roam; February 4th 2010 at 05:22 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by ADY View Post
    Ok, so how to do we use the quotient rule to show that this function

    g(x) = \frac{e^{(20-x/36)}}{(5+x)^2} ................ (x>-5)

    has derive.

     k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}

    Thank you
    u = e^{20-x/36} u' = -\frac{1}{36}e^{20-x/36}

    v = (5+x)^2 v' = 2(5+x)

    g'(x) = \frac{-(5+x)^2 \cdot \frac{1}{36}e^{20-x/36} - 2(5+x)e^{20-x/36}}{(5+x)^4} = -\frac{e^{20-x/36}\left({x} +77\right)}{36(5+x)^3}
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  4. #4
    ADY
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    looking at both your posts, they are both different solutions arent they?
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