Thread: Simplifying a Derivative as a Product of Factors: f(x) = (3x+4)^4 (2x-1)^7

1. Simplifying a Derivative as a Product of Factors: f(x) = (3x+4)^4 (2x-1)^7

Hello, I'm preparing for a math test on derivatives and I have been caught up on this type of question multiple times. I can't seem to figure out how to properly simplify the function:
Find the derivative of the function. Leave your answer as a product of factors with no negative exponents.
f(x) = (3x+4)^4 (2x-1)^7

My steps thus far have been:
f'(x) = (3x+4)^4 (7) (2x-1)^6 (2) + (2x-1)^7 (4)(3x+4)^3 (3)
f'(x) = (14) (3x+4)^4 (2x-1)^6 + (12)(2x-1)^7 (3x+4)^3
f'(x) = 2 (3x+4)^3 (2x-1)^6 [(7)(2x-1) + (6)(3x+4)]
f'(x) = 2 (3x+4)^3 (2x-1)^6 (32x +17)

Steps 3 and 4 are the most confusing for me, and I'm uncertain if I'm doing this right. I've done it this way before (for other functions similar to this) and I've gotten it wrong so I'm unsure how to approach further simplifying the function.
If you wouldn't mind explaining the steps for this if this is wrong? I've seen other students get 22(3x+4)^3 (2x-1)^6 (3x+2) and I'm not sure how they got that answer if that is correct.

Thank you so much for your time and I hope this makes sense.

2. Originally Posted by Yelhsa
… My steps thus far have been:
f'(x) = (3x+4)^4 (7) (2x-1)^6 (2) + (2x-1)^7 (4)(3x+4)^3 (3)
f'(x) = (14) (3x+4)^4 (2x-1)^6 + (12)(2x-1)^7 (3x+4)^3
f'(x) = 2 (3x+4)^3 (2x-1)^6 [(7)(2x-1) + (6)(3x+4)]
Oops, your factors 7 and 6 need to switch places.

You will then get the same result as those other students.

3. Let's try a way that cuts down on long algebraic expressions where it is easy to make a mistake.

$p = 3x + 4 \implies p' = 3.$ Easy.

$q = 2x - 1 \implies q' = 2.$ Easy.

$u = p^4 \implies u' = 4p^3 * p' = 12p^3.$ Easy.

$v = q^7 \implies v' = 7q^6 * q' = 14q^6.$ Easy

$y = uv \implies y' = uv' + u'v = u(14q^6) + u'(q^7) = q^6(14u + qu') =$

$q^6\{14p^4 + q(12p^3)\} = 2p^3q^6(7p + 6q) = 2p^3q^6\{7(3x + 4) + 6(2x - 1)\} =$

$2p^3q^6(21x + 28 + 12x - 6) = 2p^3q^6(33x + 22) = 22p^3q^6(3x + 2) =$

$22(3x + 4)^3(2x - 1)^6(3x + 2).$

Using the technique of substitution of variables is a little slower, but I have found it far less prone to error.

4. Originally Posted by mmm4444bot
Oops, your factors 7 and 6 need to switch places.

You will then get the same result as those other students.
Ah, thank you!

5. Originally Posted by JeffM
Let's try a way that cuts down on long algebraic expressions where it is easy to make a mistake.

$p = 3x + 4 \implies p' = 3.$ Easy.

$q = 2x - 1 \implies q' = 2.$ Easy.

$u = p^4 \implies u' = 4p^3 * p' = 12p^3.$ Easy.

$v = q^7 \implies v' = 7q^6 * q' = 14q^6.$ Easy

$y = uv \implies y' = uv' + u'v = u(14q^6) + u'(q^7) = q^6(14u + qu') =$

$q^6\{14p^4 + q(12p^3)\} = 2p^3q^6(7p + 6q) = 2p^3q^6\{7(3x + 4) + 6(2x - 1)\} =$

$2p^3q^6(21x + 28 + 12x - 6) = 2p^3q^6(33x + 22) = 22p^3q^6(3x + 2) =$

$22(3x + 4)^3(2x - 1)^6(3x + 2).$

Using the technique of substitution of variables is a little slower, but I have found it far less prone to error.
Ah, thank you so much!