chain rule + product rule problem: (x-3)^3(3x+7)

[oscarxue]

Hi, I've been working through this problem for ages and just couldn't find the simplified answer for it. I've checked it multiple times with a calc and couldn't get the same answer.

The problem is
(x-3)^3(3x+7)

The answer is:
12(x+1)(x-3)^2

Please show me how to simplify this problem with the use of product and chain rule.

 

[Harry_the_cat]

Hi, I've been working through this problem for ages and just couldn't find the simplified answer for it. I've checked it multiple times with a calc and couldn't get the same answer.

The problem is
(x-3)^3(3x+7)

The answer is:
12(x+1)(x-3)^2

Please show me how to simplify this problem with the use of product and chain rule.

Let u = (x-3)^3 and v = (3x+7). Find u' and v' and apply the product rule. Do this and then factorise the resulting expression.

 

[oscarxue]

Let u = (x-3)^3 and v = (3x+7). Find u' and v' and apply the product rule. Do this and then factorise the resulting expression.

Hi,

I managed to do that a while ago, ended up with
3(x-3)^2 x (3x+7) + 3(x-3)^3

What I dont get is, how do I factorise this in order to get to the correct answer?

 

[HallsofIvy]

Since you have \(\displaystyle (3x- 1)^2\) in one term and \(\displaystyle (3x- 1)^3\) in the other, an obvious first step is to factor out \(\displaystyle (3x- 1)^2\). That leaves \(\displaystyle (3x-1)^2(3(3x+7)+ 3(3x-1))\)\(\displaystyle = (3x- 1)^2(9x+ 21+ 9x- 3)= (3x- 1)^2(18x+ 18)= 18(3x- 1)^2(x+ 1)\).

 

[stapel]

Hi, I've been working through this problem for ages and just couldn't find the simplified answer for it.

What did you get. (The best way for us to help you correct your errors is for us to be able to examine them.)

The problem is
(x-3)^3(3x+7)

The answer is:
12(x+1)(x-3)^2

Did the book really say this, or was there maybe a "y" or an "f(x) for the "problem", and maybe a "dy/dx" or "f'(x)" for the "answer"?

Please show me how to simplify this problem with the use of product and chain rule.

Did the instructions really say to "simplify" (which would not involve the Product or Chain Rule), or did they maybe tell you to "differentiate" or "find the derivative of" the "problem"?

Clarity helps understanding, both for you and for the grader. And understanding boosts grades!

 

[Harry_the_cat]

Hi,

I managed to do that a while ago, ended up with
3(x-3)^2 x (3x+7) + 3(x-3)^3

What I dont get is, how do I factorise this in order to get to the correct answer?

Compare it to this:

\(\displaystyle 3a^2b +3a^3\) where a is the first bracket and b is the second.

Factorising gives;

\(\displaystyle 3a^2(b+a)\)

Same deal above gives:

\(\displaystyle 3(x-3)^2[(3x+7) +(x-3)] = 3(x-3)^2(4x+4) = 12(x-3)^2(x+1)\)