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

radnorgardens

New member
Joined
Dec 2, 2014
Messages
26
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.
 
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???
 
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.
 
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)?
 
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?

 
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?

 
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

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

\(\displaystyle \therefore f'(x) = 6x^2k^2 + 2xk^3 = 2(3x^2k^2 + xk^3) = 2x(3xk^2 + k^3) = 2xk^2(3x + k) =\)

\(\displaystyle 2xk^2(3x + 2x + 1) = 2xk^2(5x + 1) = 2x(2x + 1)^2(5x + 1) \ \because \ k = 2x + 1.\)
 
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


 
Denis virtually did

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

\(\displaystyle \therefore f'(x) = 6x^2k^2 + 2xk^3 = 2(3x^2k^2 + xk^3) = 2x(3xk^2 + k^3) = 2xk^2(3x + k) =\)

\(\displaystyle 2xk^2(3x + 2x + 1) = 2xk^2(5x + 1) = 2x(2x + 1)^2(5x + 1) \ \because \ k = 2x + 1.\)

Brilliant. Thank you for showing me this JeffM. I followed this perfectly. Just need to practice more. Many thanks to all.
 
Top