Understanding the Chain Rule: A Derivative Problem Solution

Phyzwizz
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I did a derivative problem, but my book says that my answer is wrong.

f(x)=x2(x-2)4

I didn't see much use in the chain rule so I used the product rule.

x2(4(x-2)3) + (x-2)4(2x)
=4x2(x-2)3 + 2x(x-2)4

The book says that instead of this, the answer is ...
x2(4(x-2)3(1)) + (x-2)4(2x) = 2x(x-2)3(3x-2)

So I guess where I'm really confused is in the manner with which the book simplified.
 
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Phyzwizz said:
I did a derivative problem, but my book says that my answer is wrong.

f(x)=x2(x-2)4

I didn't see much use in the chain rule so I used the product rule.

x2(4(x-2)3) + (x-2)4(2x)
=4x2(x-2)3 + 2x(x-2)4
Well, you did use the chain rule when you said that the derivative of (x- 2)4 is 4(x- 2)3 (because the derivative of x- 2 is 1).

If you factor out 2x(x- 2)3 to get f'= 2x(x- 2)3(2x+ (x- 2))= 2x(x- 2)(3x- 2), exactly what your book says. Your answer and your book's answer are the same.

The book says that instead of this, the answer is ...
x2(4(x-2)3(1)) + (x-2)4(2x) = 2x(x-2)3(3x-2)

So I guess where I'm really confused is in the manner with which the book simplified.
 
Awesome thanks, I get it now. And I also now understand how ridiculous my book is in its mindless factoring.
 

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