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Math Help - Product Rule

  1. #1
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    Product Rule

    I'm new to all this stuff so bear with me...

    I'm given  y =2x(8x+6)

    I need to find  y'

    So i do it...

     \frac {dx}{dy} = u\frac{dv}{dx} + v\frac {du}{dx}

     2x.\frac{d(8x+6)}{dx} + (8x+6).\frac {d2x}{dx}

     16x+12 + 16x +12

     32x +24

    But when I check on my calculator it says  32x+12

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  2. #2
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    Quote Originally Posted by jgv115 View Post
    I'm new to all this stuff so bear with me...

    I'm given  y =2x(8x+6)

    I need to find  y'

    So i do it...

     \frac {dx}{dy} = u\frac{dv}{dx} + v\frac {du}{dx}

     2x.\frac{d(8x+6)}{dx} + (8x+6).\frac {d2x}{dx}

     16x+12 + 16x +12

     32x +24

    But when I check on my calculator it says  32x+12

    y = 2x(8x+6).


    \frac{dy}{dx} = u\,\frac{dv}{dx} + v\,\frac{du}{dx}

     = 2x(8) + (8x + 6)(2)

     = 16x + 16x + 12

     = 32x + 12.
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  3. #3
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    Hey that's exactly what my book says! But I always get different.

    = 2x(8) + (8x + 6)(2)

    I don't understand how you get  2x(8)

    How can \frac{d(8x+6)}{dx} become 8?
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    Quote Originally Posted by jgv115 View Post
    Hey that's exactly what my book says! But I always get different.

    = 2x(8) + (8x + 6)(2)

    I don't understand how you get  2x(8)

    How can \frac{d(8x+6)}{dx} become 8?
    The derivative of 8x is 8 and the derivative of 6 is 0.

    So the derivative of 8x + 6 is 8 + 0 = 8.
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    oohh!!!!!!!! Thanks so much!!!!!!!

    If I have  \frac {d(7x+6)}{dx}

    Then would it be  7 ??
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  6. #6
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    Quote Originally Posted by jgv115 View Post
    oohh!!!!!!!! Thanks so much!!!!!!!

    If I have  \frac {d(7x+6)}{dx}

    Then would it be  7 ??
    Correct.
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  7. #7
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    Thanks Prove It!! I appreciate the help!
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