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

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

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

4. Originally Posted by radnorgardens
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. Originally Posted by lev888
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. Originally Posted by radnorgardens
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. Originally Posted by Denis
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. Originally Posted by lev888
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. Instead of a*(b+c) you got a+b+c.

10. Originally Posted by lev888
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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