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Thread: Simplification of Product Rule: dy/dx = x^2 [6(2x + 1)^2] + (2x + 1)^3 (2x)

  1. #11
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    Quote Originally Posted by radnorgardens View Post
    Ok. I'm not getting this. Using a*(b+c) how does this relate to the textbook solution?

    dy/dx = x2[6(2x + 1)2] + (2x + 1)3(2x)

    Common factor = 2x(2x + 1)2

    dy/dx = 2x(2x + 1)2 [3x + (2x + 1)]
    dy/dx = 2x(2x + 1)2 (5x + 1)]

    Is it possible someone can show my step by step?

    Denis virtually did

    [tex]f'(x) = x^2[6(2x + 1)^2] + (2x + 1)^3(2x) = 6x^2k^2 + 2xk^3, \text { where } k = 2x + 1.[/tex]

    [tex]\therefore f'(x) = 6x^2k^2 + 2xk^3 = 2(3x^2k^2 + xk^3) = 2x(3xk^2 + k^3) = 2xk^2(3x + k) =[/tex]

    [tex]2xk^2(3x + 2x + 1) = 2xk^2(5x + 1) = 2x(2x + 1)^2(5x + 1) \ \because \ k = 2x + 1.[/tex]

  2. #12
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    Quote Originally Posted by radnorgardens View Post
    Ok. I'm not getting this. Using a*(b+c) how does this relate to the textbook solution?
    Ok, let's try a simpler example. Please use the rule a*b + a*c = a*(b+c) to write the following expression as a product:
    4x + 3x2



  3. #13
    Quote Originally Posted by JeffM View Post
    Denis virtually did

    [tex]f'(x) = x^2[6(2x + 1)^2] + (2x + 1)^3(2x) = 6x^2k^2 + 2xk^3, \text { where } k = 2x + 1.[/tex]

    [tex]\therefore f'(x) = 6x^2k^2 + 2xk^3 = 2(3x^2k^2 + xk^3) = 2x(3xk^2 + k^3) = 2xk^2(3x + k) =[/tex]

    [tex]2xk^2(3x + 2x + 1) = 2xk^2(5x + 1) = 2x(2x + 1)^2(5x + 1) \ \because \ k = 2x + 1.[/tex]
    Brilliant. Thank you for showing me this JeffM. I followed this perfectly. Just need to practice more. Many thanks to all.

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