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Thread: derivatives

  1. #1
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    derivatives

    find dy/dx for
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  2. #2
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    Quote Originally Posted by algebra2 View Post
    find dy/dx for
    First of all, remember that a square root is a power of $\displaystyle \frac{1}{2}$.

    You have a product of functions, so use the product rule.

    $\displaystyle \frac{dy}{dx} = (2x+1)^{\frac{1}{2}}\times \frac{d}{dx} (x^3) + x^3 \times \frac{d}{dx}[(2x+1)^{\frac{1}{2}}]$.

    To work out the second of the derivatives, use the chain rule.
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    $\displaystyle (2x+1)^{\frac{1}{2}}\times (3x^2)
    $



    for here, do i multiply the 3x^2 into the (2x+1)^(1/2) to get (6x+3x)^(5/2)?
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    Quote Originally Posted by algebra2 View Post
    $\displaystyle (2x+1)^{\frac{1}{2}}\times (3x^2)$$\displaystyle
    $


    for here, do i multiply the 3x^2 into the (2x+1)^(1/2) to get (6x+3x)^(5/2)?
    No, just write it as $\displaystyle 3x^2(2x+1)^{\frac{1}{2}}$.
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