How does the product and chain rule apply to this problem?

[JoshHolloway]

Could someone please help me, I do not understand how the author of my textbook gets from one point to another. Here is the problem worked out, after the problem I will explain which part I don't understand.

f(x)=x(x-4)^3
f'(x)=x[3(x-4)^2]+(x-4)^3
=(x-4)^2(4x-4)

I do not understand how the author gets from here:
f'(x)=x[3(x-4)^2]+(x-4)^3
to here:
=(x-4)^2(4x-4)

 

[JoshHolloway]

When I do it it looks like this:
f'(x)=x[3(x-4)^2]+(x-4)^3
=x[3(x^2-8x+16)]+(x-4)(x^2-8x+16)
=x(3x^2-24x+48)+(x^3-8x^2+16x-4x^2+32x-64)
=3x^3-24x^2+48x+x^3-8x^2+16x-4x^2+32x-64
=4x^3-36x^2+96x-64
=4(x^3-9x^2+24x-16)

 

[JoshHolloway]

I am obviously doing something wrong, but I can't seem to figure it out. It will help me a great deal when I figure out what I am doing wrong in these types of problems, so any help is greatly apreciated.

 

[jma2001]

I believe it was done by factoring (x-4)^2 out of the second line.

So, you end up with: (x-4)^2[3x+(x-4)] which reduces to the final expression.

 

[JoshHolloway]

I don not understand how you can factor (x-4)^2 out of the second line.

 

[jma2001]

The square brackets in the second line are redundant and may be part of what is confusing you. That is:

x[3(x-4)^2] = 3x(x-4)^2

Now, rewrite the second line and see if it makes more sense.

 

[JoshHolloway]

Jimminy Christmas!
It's that easy. Man, I can't believe that. I always mess stupid things up like that. Now it is time for me to find the second derivative.

 

[jma2001]

I'm glad you understand it. That sort of thing used to confuse me all the time, and it still does on occasion. The only way to get better at "seeing" those quick solutions is with lots of practice.

 

[JoshHolloway]

I am learning that the hard way jma. By the way, thanks for the help, I would have never seen that without the assistance.