# Derivative of [(1+4x)^5][(3+x-x^2)^8]

1. Jul 26, 2008

### illjazz

1. The problem statement, all variables and given/known data
Find the derivative of the function

$$g(x)=(1+4x)^5(3+x-x^2)^8$$

2. Relevant equations
- Chain rule
- Power rule

3. The attempt at a solution

$$g(x)=(1+4x)^5(3+x-x^2)^8$$

$$g(x)=(1+4x)^5\frac{d}{dx}(3+x-x^2)^8+(3+x-x^2)^8\frac{d}{dx}(1+4x)^5$$

$$g(x)=(1+4x)^58(2x-1)(3+x-x^2)^7+(3+x-x^2)^820(1+4x)^4$$
...

This is where I'm stuck. Both my calculator and the book tell me that the solution is

$$4(4x+1)^4(x^2-x-3)^7(21x^2-9x-17)$$

All I want to know is how they get to this result from the step I got stuck at above. I don't see how they arrive at that last polynomial :/

Last edited: Jul 26, 2008
2. Jul 26, 2008

### rocomath

I don't understand how you got to your 3rd step. You didn't bring down your power 8 for the first part and you have a 0 on the second part.

$$\frac{d}{dx}(3+x-x^2)^8=8(3+x-x^2)^7(1-2x)=8(2x-1)(x^2-x-3)^7$$

$$\frac{d}{dx}(1+4x)^5=5(1+4x)^4(4)=20(1+4x)^5$$

3. Jul 26, 2008

### illjazz

My bad.. I'm still learning this tex stuff. Inserting spaces using "\" just causes whatever character that follows to disappear completely.. and adding new lines doesn't work at all (with "\\"). Painful.

I used spaces after the exponents because I thought if I wrote an "8" right after "5", I'd get an exponent of 58.. instead, my 8 disappeared completely. Bleh.. turns out you can just write x^58(x+1) in tex and the result will be (x^5)(8(x+1)).

Anyway...

4. Jul 26, 2008

### rocomath

Your signs are still messed up. You wrote (2x-1)(3+x-x^2)^7.

You need to put a negative out in the front or factor out another one from the (3+x-x^2)^7 term.

5. Jul 26, 2008

### illjazz

$$g(x)=(1+4x)^58(1-2x)(3+x-x^2)^7+(3+x-x^2)^820(1+4x)^4$$
Is this what it's supposed to be? Because

$$\frac{d}{dx}(3+x-x^2)^8=8(3+x-x^2)^7*(1-2x)$$

6. Jul 26, 2008

### rocomath

Yes that's good now. I guess you'll need to just continue simplifying so it matches the book's answer unless your teacher doesn't care if you don't simplify further.

7. Jul 26, 2008

### illjazz

$$g(x)=(1+4x)^5(3+x-x^2)^8$$

$$g'(x)=(1+4x)^5\frac{d}{dx}(3+x-x^2)^8+(3+x-x^2)^8\frac{d}{dx}(1+4x)^5$$

$$g'(x)=(1+4x)^58(3+x-x^2)^7*\frac{d}{dx}(3+x-x^2)+(3+x-x^2)^85(1+4x)^4*\frac{d}{dx}(1+4x)$$

$$g'(x)=(1+4x)^58(3+x-x^2)^7(1-2x)+(3+x-x^2)^85(1+4x)^4*4$$

$$g'(x)=8(1+4x)^5(3+x-x^2)^7(1-2x)+20(3+x-x^2)^8(1+4x)^4$$

Let (1+4x) = u
Let (3+x-x^2) = v
Let (1-2x) = w

Then

$$g'(x)=8u^5v^7w+20v^8u^4$$

$$g'(x)=4u^4v^7(2uw+5v)$$

So

$$g'(x)=4(1+4x)^4(3+x-x^2)^7(2(1+4x)(1-2x)+5(3+x-x^2))$$

$$g'(x)=4(1+4x)^4(3+x-x^2)^7(2(1+2x-8x^2)+15+15x-5x^2)$$

$$g'(x)=4(1+4x)^4(3+x-x^2)^7(2+4x-16x^2+15+15x-5x^2)$$

$${g'(x)=4(1+4x)^4(3+x-x^2)^7(17+19x-21x^2)$$

And that final line is what my book has as the answer. Phew.. that substitution definitely helped the thinking process. Also, I realized one mistake I was making with the Chain Rule.

Say we have

$$y = (x+3)^3$$

Then I would do

$$y' = 3(x+3)^2*\frac{d}{dx}(x+3)^3$$

$$y' = 3(x+3)^2*\frac{d}{dx}(x+3)$$...