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Math Help - Product rule. (checking)

  1. #1
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    Product rule. (checking)

    one more guys please

    differentiate using the prduct rule

     <br />
y(x) = (5x^3 + 8x^2 - 2x + 1)e^x<br />

    <br />
u = (5x^3 + 8x^2 - 2x + 1) v = e^x<br />

    <br />
\frac{du}{dx}= (10x^2 + 16x - 2)<br />

    so

    <br />
\frac{dy}{dx}= v\frac{du}{dx} + u\frac{dv}{dx} = e^x(10x^2 + 16x - 2) + (5x^3 + 8x^2 - 2x + 1)e^x<br />
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  2. #2
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    Quote Originally Posted by mxmadman_44 View Post
    one more guys please

    differentiate using the prduct rule

     <br />
y(x) = (5x^3 + 8x^2 - 2x + 1)e^x<br />
    What are you doing?

    y'=(5x^3+8x^2-2x+1)'e^x+(5x^3+8x^2-2x+1)(e^x)'
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  3. #3
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    .

    could you please show you workings how you gotto that answer i dont understand how you got v and u. I take it there is no need to differentiate anything. ie
    <br />
3x^2  = 6x<br />
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  4. #4
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    The product rule states that:
    (uv)' = u'v+uv'

    Here we let u=5x^3+8x^2-2x+1 \mbox{ and }v=e^x
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  5. #5
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    ..

    okay got it, thanks!. I had a different equation but your answer looks right. Plus your know way more than me.

    Cheers perfecthacker
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