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

[illjazz]

Homework Statement

Find the derivative of the function

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

Homework Equations

- Chain rule
- Power rule

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 :/

 

[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.

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

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

 

[illjazz]

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.

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

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

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...

 

[rocomath]

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

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.

 

[illjazz]

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.

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

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

 

[rocomath]

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

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

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.

 

[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

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

instead of

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

At least I think that's a mistake I used to make. Now I know :). How the heck do you make stuff bold inside tex expressions? And I'd still love to know why the "\" and "\\" escape characters do not produce what they should when using TeX/LaTeX on physicsforums.com.

That's my long, drawn out solution to the problem.. but hey, it helped and I felt like doing it all out :)