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Thread: Finding the derivative using product rule

  1. #1
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    Finding the derivative using product rule

    Expand x(2x+1)^4, then find the derivative.

    How would I find the derivative?
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  2. #2
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    Re: Finding the derivative using product rule

    Quote Originally Posted by EarlyTravel View Post
    Expand x(2x+1)^4, then find the derivative.

    How would I find the derivative?
    You have 2 functions, $\displaystyle f(x) $ and $\displaystyle g(x) $. Let's say $\displaystyle f(x) = x $ and $\displaystyle g(x) = (2x+1)^4 $.

    The product rule is $\displaystyle f'(x)g(x) + g'(x)f(x) $

    1) $\displaystyle (\frac{d}{dx}x)(2x+1)^4 + (\frac{d}{dx}(2x+1)^4)x $. Notice that $\displaystyle (\frac{d}{dx}(2x+1)^4) $ requires the Chain Rule.


    The Chain Rule is $\displaystyle f(g(x))' = f'(g(x))*g'(x) $ , so let's say $\displaystyle f(x) = x^4 $ and $\displaystyle g(x) = 2x +1 $ then $\displaystyle (4(2x+1)^3(2)) $

    Back to the original problem:

    2) $\displaystyle 1(2x+1)^4 + 4(2x+1)^3(2))x $.

    3) $\displaystyle (2x+1)^4 + 8x(2x+1)^3 $.
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  3. #3
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    Re: Finding the derivative using product rule

    Quote Originally Posted by AZach View Post
    Back to the original problem:

    2) $\displaystyle 1(2x+1)^4 + 4(2x+1)^3(2))x $.

    3) $\displaystyle (2x+1)^4 + 8x(2x+1)^3 $.
    one more worthwhile step ...

    $\displaystyle (2x+1)^4 + 8x(2x+1)^3 = (2x+1)^3[(2x+1) + 8x] = (2x+1)^3(10x+1)$

    ... makes finding critical points a bit less work.
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    Re: Finding the derivative using product rule

    Hello, EarlyTravel!

    Expand $\displaystyle f(x) \:=\:x(2x+1)^4$, then find the derivative.

    How would I find the derivative? . ??

    I don't understand your question.
    . . You don't know how to expand the function?
    . . You don't know how to differentiate a polynomial?
    . . You don't know the Product Rule?


    How about following the directions?

    Expand: .$\displaystyle f(x) \:=\:x(2x+1)^4 $

    . . . . . . . . . . . $\displaystyle =\;x(16x^4 + 32x^3 + 24x^2 + 8x + 1)$

    . . . . . . . . . . . $\displaystyle =\;16x^5 + 32x^4 + 24x^3 + 8x^2 + x$

    Differentiate: .$\displaystyle f'(x) \;=\;80x^4 + 128x^3 + 72x^2 + 16x + 1$ .[1]



    Product Rule:

    .$\displaystyle f(x) \:=\:\overbrace{x}^{f(x)}\cdot\overbrace{(2x+1)^4} ^{g(x)}$

    $\displaystyle f'(x) \;=\;\overbrace{1}^{f'(x)}\cdot\overbrace{(2x+1)^4 }^{g(x)} + \overbrace{x}^{f(x)}\cdot\overbrace{4(2x+1)^3\cdot 2}^{g'(x)} $

    . . . . $\displaystyle =\;(2x+1)^4 + 8x(2x+1)^3$

    . . . . $\displaystyle =\;(2x+1)^3\big[(2x+1) + 8x\big]$

    . . . . $\displaystyle =\;(2x+1)^3(10x+1)$

    which is equal to [1].
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