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Math Help - Chain Rule

  1. #1
    Junior Member R3ap3r's Avatar
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    Chain Rule

    Is the derivative of y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(4x + 1)\Longrightarrow\; y' = 24x + 6
    Or did I screw up?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by R3ap3r View Post
    Is the derivative of y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(4x + 1)\Longrightarrow\; y' = 24x + 6
    Or did I screw up?
    Not quite

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

    using the chain rule we get

     <br />
\frac{dy}{dx}=\underbrace{6(2x^2+x)}_{PowerRule} \cdot \underbrace{4x+1}_{derivative Inside}<br />
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  3. #3
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    Sadly, I think you screwed up. When you brought the second power down, remember that you're just reducing the power by one, i.e., the parenthetical 2x^2 +  x is still being raised to the first power.

    y = 3(2x^2 + x)^2\Longrightarrow\; y' = 6(2x^2 + x)(4x + 1)\Longrightarrow\; y' = 6(8x^3 + 6x^2 + x)

    so

    y'=48x^3 + 36x^2 + 6x.

    Note your original equation is a fourth-degree polynomial (since it has an x-squared term squared), so your derivative should be a third degree polynomial.
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  4. #4
    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Not quite

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

    using the chain rule we get

     \frac{dy}{dx}=\underbrace{6(2x^2+x)}_{PowerRule} \cdot \underbrace{4x+1}_{derivative Inside}<br />
    So your saying multiply the 6(2x^2+x) out then multiply that by (4x+1) then theres the derivative?
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  5. #5
    Senior Member topher0805's Avatar
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    Quote Originally Posted by R3ap3r View Post
    So your saying multiply the 6(2x^2+x) out then multiply that by (4x+1) then theres the derivative?
    When using the chain rule you follow this general formula:

    The derivative of f(g(x)) is equal to f'(g(x))\cdot g'(x).
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  6. #6
    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by topher0805 View Post
    When using the chain rule you follow this general formula:

    The derivative of f(g(x)) is equal to f'(g(x))\cdot g'(x).
    Alright so the poster above me is correct then.
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by R3ap3r View Post
    So your saying multiply the 6(2x^2+x) out then multiply that by (4x+1) then theres the derivative?
    yes let

     f(u)=3u^2 \mbox{ and } f'(u)=6u

    and

     u(x)=2x^2+x \mbox{ and } u'(x)=4x+1

    so

     y=f(u(x))=3(2x^2+x)^2

    so by defintion of the chain rule we get

     y'= f'(u(x)) \cdot u'(x)=6(2x^2+x) \cdot (4x+1)

    I hope this helps.
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  8. #8
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    You can multiply the parts of the derivative in whatever order you want - multiplication is commutative and can be performed in any order. I chose to FOIL multiply the two binomials together (combining the outer and inner terms, as they were both x^2 terms) and then lastly distributing the 6 through.
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