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Math Help - Calculus problem

  1. #1
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    Calculus problem

    I am not sure if I am on the right track with this problem and where do i go from here. Please help.

    Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

    -8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    -112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    Where do I go from here?

    Thank you in advance. Any help is greatly appreciated.

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  2. #2
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    Quote Originally Posted by Azathoth View Post
    I am not sure if I am on the right track with this problem and where do i go from here. Please help.

    Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

    -8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    -112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    Where do I go from here?

    Thank you in advance. Any help is greatly appreciated.
    Use the product rule (quotient rule is possible here too):

    f(x)=4x^2 \cdot (7x^2+8)^{-\frac23}

    f'(x)= (7x^2+8)^{-\frac23} \cdot 8x + 4x^2 \cdot \left(-\frac23\right) (7x^2+8)^{-\frac53} \cdot 14x

    Transform into a quotient, the denominator is 3(7x^2+8)^{-\frac53}

    f'(x)=\dfrac{3 \cdot 8x((7x^2+8) - 2 \cdot 4x^2 \cdot 14x}{3(7x^2+8)^{\frac53}} = \dfrac{8x \cdot (7x^2+24)}{3(7x^2+8)^{\frac53}}
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  3. #3
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    So was I on the wrong track or was I just going about it in a longer version? So the answer is [IMG]file:///C:/Users/Michelle/AppData/Local/Temp/moz-screenshot.jpg[/IMG]8x(7x^2 +24) over 3(7x^2 +8)^(5/3)? If so, eventually would I have gotten to that answer on the path I was taking? Thank you for your help.
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  4. #4
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    Quote Originally Posted by Azathoth View Post
    ...

    Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

    -8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    -112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
    ...
    Quote Originally Posted by Azathoth View Post
    So was I on the wrong track or was I just going about it in a longer version?

    ...
    I've got exactly the same result as you did. I only simplified the result using a convenient common denominator. Nothing else.
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