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Math Help - 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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    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,  f(x) and  g(x) . Let's say  f(x) = x and  g(x) = (2x+1)^4 .

    The product rule is  f'(x)g(x) + g'(x)f(x)

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


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

    Back to the original problem:

    2)  1(2x+1)^4 + 4(2x+1)^3(2))x .

    3)  (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)  1(2x+1)^4 + 4(2x+1)^3(2))x .

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

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

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

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

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



    Product Rule:

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

    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)}

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

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

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

    which is equal to [1].
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