# Thread: Confusing Algebra in a Derivative

1. ## Confusing Algebra in a Derivative

Hi, I couldn't find anyone around my dorm that could help me so I thought I would try a forum. Thanks for any help I receive!

My problem is:

Find the derivative:
y = (3x^4 + 1)^4 * (x^3 + 4)

According to my book the answer is 3x^2(3x^4 + 1)^3 * (19x^4 + 64x + 1)

For some reason I never can get this solution.

Here are the 2 steps in the book that really confuse me:
3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3
to this
3x^2(3x^4 + 1)^3[3x^4 + 1 + 16x(x^3 + 4)]

I get to the first step but I do not get the second step. I think I am just overlooking something simple, but I don't see it.

Thanks!!

2. y = (3x^4 + 1)^4 * (x^3 + 4)

To find the derivative, we're going to need to use a chain rule and a product rule.

y'= (3x^4 +1)^4 * d/dx (x^3 +4) + (x^3+4) * d/dx (3x^4 +1)^4
= (3x^4 +1)^4 * 3x^2 + (x^3+4) * 4(3x^4 +1)^3 * 12x
= 3x^2(3x^4 +1)^3 + 48x(x^3+4)(3x^4 +1)^3

From here, all you need to do is simplify the expression to get the textbook answer.

Oh, I see, didn't read your question thoroughly enough right away. Um, all you really should have to do is simplify the expression and then factor out a 3x^2(3x^4 + 1)^3 from the expression.

3. Originally Posted by stunirvana
Hi, I couldn't find anyone around my dorm that could help me so I thought I would try a forum. Thanks for any help I receive!

My problem is:

Find the derivative:
y = (3x^4 + 1)^4 * (x^3 + 4)

According to my book the answer is 3x^2(3x^4 + 1)^3 * (19x^4 + 64x + 1)

For some reason I never can get this solution.

Here are the 2 steps in the book that really confuse me:
3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3
to this
3x^2(3x^4 + 1)^3[3x^4 + 1 + 16x(x^3 + 4)]

I get to the first step but I do not get the second step. I think I am just overlooking something simple, but I don't see it.

Thanks!!
$\displaystyle 3x^2(3x^4 + 1)^4 + 48x^3(x^3 + 4)(3x^4 + 1)^3$

$\displaystyle \textcolor{red}{3x^2(3x^4 + 1)^3} \cdot (3x^4+1) + \textcolor{red}{3x^2} \cdot 16x(x^3 + 4)\textcolor{red}{(3x^4 + 1)^3}$

note what factors the two terms have in common ...

both have the common factors $\displaystyle \textcolor{red}{3x^2}$ and $\displaystyle \textcolor{red}{(3x^4+1)^3}$ ... they are both factored out from each term ...

$\displaystyle \textcolor{red}{3x^2(3x^4+1)^3}[(3x^4+1) + 16x(x^3+4)]$

any clearer now?

4. I can get this. I think it should be:
Originally Posted by Math Major
y = (3x^4 + 1)^4 * (x^3 + 4)
y'= (3x^4 +1)^4 * d/dx (x^3 +4) + (x^3+4) * d/dx (3x^4 +1)^4
= (3x^4 +1)^4 * 3x^2 + (x^3+4) * 4(3x^4 +1)^3 * 12x^3
= 3x^2(3x^4 +1)^4 + 48x^3(x^3+4)(3x^4 +1)^3
Correct me if I am wrong above.

Its the simplification I'm having trouble with.

5. Originally Posted by skeeter

any clearer now?
So I just had to pull a $\displaystyle (3x^4+1)$ out of $\displaystyle (3x^+1)^4$ then pull them (I mean the two terms you highlighted) out to the front?

By the way, the math text thing is awesome!

6. Yes, you're correct. I just typed it too quickly.