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

  1. #1

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

    Hi all,

    I'm learning the rules of differentiation, specifically the product rule. I understand the steps to obtain an answer, but am stuck when it comes to simplifying and taking out a common factor. For example, the answer to one of the problems is:

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

    I understand how to obtain this answer, but not the simplification. Continuing.....take out a common factor 2x(2x + 1)2.

    Apparently, this goes into the first term 3x times. And the second term 2x + 1 times. Resulting in:

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

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


    As always, thanks for any help. I'd appreciate if someone could show me, step by step.

    B.

  2. #2
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    Do you see 2x+1 in both terms?

  3. #3
    Quote Originally Posted by lev888 View Post
    Do you see 2x+1 in both terms?
    Yes....it's the 3x and 2x outside the brackets that is most confusing.

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    Quote Originally Posted by radnorgardens View Post
    Yes....it's the 3x and 2x outside the brackets that is most confusing.

    3x is what's left from the first term after division by the common factor. 2x+1 is what's left from the second term.

  5. #5
    Quote Originally Posted by lev888 View Post
    3x is what's left from the first term after division by the common factor. 2x+1 is what's left from the second term.
    dy/dx = x2[6(2x + 1)2] + (2x + 1)3(2x)
    becomes
    dy/dx = 6x2[(2x + 1)2] + (2x)(2x + 1)3

    Ok, I see 2x(2x + 1)2 goes into the 2nd term, leaving (2x + 1)

    So, now
    2x(2x + 1)2 into the 1st term of 6x2[(2x + 1)2] . I'm not following here....I know this is going to be wrong, but just so you can see my flawed thinking:

    Maybe transform 1st term to (x)
    6x[(2x + 1)2]. Then 6x[(2x + 1)2] divided by 2x(2x + 1)2 goes in three times, just leaving 3.

    So (x) 3 + (2x + 1). i.e. 3x + (2x + 1) or 5x + 1.

    Why in my textbook answer is there still another
    2x(2x + 1)2???

  6. #6
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    Quote Originally Posted by radnorgardens View Post
    Why in my textbook answer is there still another 2x(2x + 1)2???
    This is the common factor. You can't just divide by it and throw it away - the value of the expression would change. Do you remember this formula: a*b + a*c = a*(b+c)? This is "a", the common factor.

  7. #7
    Quote Originally Posted by Denis View Post
    Try this: let k = 2x + 1
    Then:
    dy/dx = x^2(6k^2) + (k^3)2x
    dy/dx = 6x^2(k^2) + 2x(k^3)
    dy/dx = k^2(6x^2 + 2xk)

    ...OK?

    Ok. I follow that Denis. But how to get the textbook answer (so I can understand his process)?

  8. #8
    Quote Originally Posted by lev888 View Post
    This is the common factor. You can't just divide by it and throw it away - the value of the expression would change. Do you remember this formula: a*b + a*c = a*(b+c)? This is "a", the common factor.
    I remember, so to revise:

    1st term to (x)6x[(2x + 1)2].
    Then
    6x[(2x + 1)2] divided by 2x(2x + 1)2 goes in three times, just leaving 2x(2x + 1)2 + 3x
    So 2x(2x + 1)2 + 3x+ (2x + 1).
    Thus 2x(2x + 1)2 + 5(x + 1).

    Correct process?


  9. #9
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    Instead of a*(b+c) you got a+b+c.

  10. #10
    Quote Originally Posted by lev888 View Post
    Instead of a*(b+c) you got a+b+c.
    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?


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