Solving for f '(x) using the chain rule

[msc8127]

Homework Statement

f(x)= x^2(x-2)^4 solve for f '(x)

Homework Equations

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

The Attempt at a Solution

4x^2(x-2)^3

The answer is given in the book as 2x(x-2)^3(3x-2)

i'm not following any progression that gets me to that solution regardless of how many times I work through it.

If someone could give me a step by step on this on I'd greatly appreciate it.

Thanks
SC

 

[gabbagabbahey]

You'll need to use the product rule since f(x) is the product of x^2 and (x-2)^4.

 

[sara_87]

If you have a function of two products:
f(x)=g(x)h(x) where g and h are two functions with respect to x, you have to use the product rule so,
f '(x)= g(x)h'(x) + h(x)g'(x)
can you apply this to your function f(x)?
(in your case, g(x)=x^2 and h(x)=(x-2)^4 but it could be the other way round too)

 

[msc8127]

ok, if I'm using the product rule correctly, that takes me to:

x^2[4(x-2)^3(1)] + (x-2)^4(2x)

for some reason I'm still not seeing how that will go to 2x(x-2)^3(3x-2) for a final solution

Thanks for the help!

 

[msc8127]

sara's reply wasn't showing when I posted my last reply...let me work on it some more from what she posted and see if I can get anywhere yet.

Thanks

 

[gabbagabbahey]

ok, if I'm using the product rule correctly, that takes me to:

x^2[4(x-2)^3(1)] + (x-2)^4(2x)

for some reason I'm still not seeing how that will go to 2x(x-2)^3(3x-2) for a final solution

Thanks for the help!

Your solution is correct, to get it into the same form as the one in your answer key, factor out a 2x(x-2)^3, that gives 2x(x-2)^3[2x+(x-2)] then just simplify [2x+(x-2)]

 

[msc8127]

gotta love how algebra mistakes cause more errors than actual cal fundamentals.

I was out of school many years, in fact the last math class I had was college algebra in 2001. So, I'm really struggling with stuff that isn't all that difficult in the grand scheme of things.

Thank you both for being patient and helping me out!

 

[msc8127]

ok, one more problem here that I've tried to work. I think I'm ok on it except for the factoring out in the last step.

the given problem is f(x)= x(3x-7)^3

I got f '(x) = x[3(3x-7)^2] + (1)(3x-7)^3 via the product rule.

after attempting to factor I ended up with f '(x) = (6x-7)(3x-7)^2, which I don't think is correctly factored.

 

[sara_87]

I think you made a small mistake, when you take (3x-7)^2 as common factor, you should get:
(3x-7)^2(9x+3x-7)=(3x-7)^2(12x+7)

 

[msc8127]

once again, got to love algebra.

Thanks Sara!

 

[sara_87]

:)
U're welcome.

 

[gabbagabbahey]

The derivative of 3x-7 is (3x), not just x. You should have;

f '(x) = (3x)[3(3x-7)^2] + (1)(3x-7)^3=(12x-7)(3x-7)^2

 

[sara_87]

i meant -7
sorry. (misprint)