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Math Help - Chain rule help

  1. #1
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    Chain rule help

    y = sqrt(2x+1)(x^3)

    I always screw this up, so if someone could just walk through the steps so I could compare, I'd really appreciate it.

    Thanks.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Is it   y = x^{3} \sqrt{2x+1} ?

    Then  \frac{dy}{dx} = x^{3} \frac{1}{2}(2x+1)^{-\frac{1}{2}} + 3x^{2} \sqrt{2x+1}
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  3. #3
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    Yeah, that's what I had.

    Is it okay to leave it like that or do I have to simplify it more?

    Thanks.
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  4. #4
    Senior Member tukeywilliams's Avatar
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    you can just leave it like that.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    Is it   y = x^{3} \sqrt{2x+1} ?

    Then  \frac{dy}{dx} = x^{3} \frac{1}{2}(2x+1)^{-\frac{1}{2}} + 3x^{2} \sqrt{2x+1}
    Quote Originally Posted by alreadyinuse View Post
    Yeah, that's what I had.

    Is it okay to leave it like that or do I have to simplify it more?

    Thanks.
    You can leave it, but technically you have a radical in the denominator:
     \frac{dy}{dx} = x^{3} \frac{1}{2\sqrt{2x + 1}} + 3x^{2} \sqrt{2x+1}

    So
     \frac{dy}{dx} = \frac{x^3 \sqrt{2x + 1}}{4x + 2} + 3x^{2} \sqrt{2x+1}

    It really depends on how you are going to use the derivative from here on.

    -Dan
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  6. #6
    Senior Member tukeywilliams's Avatar
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    Or you can write it as:

     \frac{x^{3}}{2\sqrt{2x+1}} + 3x^{2} \sqrt{2x+1}
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