Simple Calc 1 question, having a block for some reason. question and work in second post.

Question

anonymous · Answer

$$f(x)=x^2(x-2)^4$$
find the derivative
use product rule
$$2x(x-2)^4+x^2(4(x-2)^3(1))$$
this is silly but this is the step i'm stuck on. I remember something about the x^2 but I cant remember why I take it out or what to do with it. any hints. please don't provide solutions. I just wanna get past my little bump so I can solve it

shamil98 · Answer

Product and chain rule, i believe is required.

anonymous · Answer

shamils got it '-'. hes alreadt here

shamil98 · Answer

Do you want to re-attempt it? Or should I give the step-by-step derivative?

anonymous · Answer

talking to me or lonnie?

anonymous · Answer

well I don't want the answer. but I can apply the chain rule to the first part. but what do I do with the x^2

shamil98 · Answer

$$x^2 (x-2)^4$$
$$2x(x-2)^4 + 4x^2(x-2)^3*1$$
well actually you've done it correctly now that i've worked it out..

anonymous · Answer

when I apply the chain rule to the first part I get 
$$8(x-2)^3(1)+x^2(4(x-2)^3)$$
something about the x^2 is making me wonder about something.. I cant remember.

shamil98 · Answer

The product and chain rule combined is:
$$\large f(x) * g(h(x))$$
$$\large f'(x)*g(h(x)) + f(x) * g'(h(x))*h'(x)$$
You did it correctly, nice job.

shamil98 · Answer

Where did you get the 8 from?

shamil98 · Answer

all you do to x^2(4(x-2)^3))
is multiply the 4 to x^2

4x^2(x-2)^3

anonymous · Answer

there has to be another step the solution shows to be
$$2x(x-2)^3(3x-2)$$
just ignore that 8. its not correct that's just me screwing around

shamil98 · Answer

$$f(x) = x^2 $$
$$g(h(x)) =(x-2)^4$$
$$f'(x) = 2x$$
$$g'(h(x))= 4(x-2)^3$$
$$h(x) = 1$$

shamil98 · Answer

Wolfram makes mistakes sometimes..

shamil98 · Answer

$$2x(x-2)^4 + x^2*4(x-2)^3 * 1$$
$$2x(x-2)^4 + 4x^2(x-2)^3$$

shamil98 · Answer

h'(x) = 1 **
small typo there.

anonymous · Answer

it was wolfram and the book. is there a rule that requires to pull the x^2 out in front of the derivative or something? I see it in my book but I don't understand why its doing it 
if you factor out the 2x there you get 
$$(2x)((x-2)^4+2(x-2)^3)$$

shamil98 · Answer

you made a typo i assume
it would be
$$(2x)((x-2)^4 + 2x(x-2)^3)$$

shamil98 · Answer

And no there is no rule for that.

anonymous · Answer

ya my bad. now that makes it close to the book answer.

shamil98 · Answer

I think the book is wrong, because using the product and chain rule combined, you should arrive at the answer...
$$2x(x-2)^4 + 4x^2(x-2)^3$$

shamil98 · Answer

No idea how it got this xD
$$2x(x-2)^3(3x-2)$$

anonymous · Answer

Alright, I know the book could be wrong. I'm just a little obsessive about problems I can't solve. I'll leave it at that, thanks for the help. Yea I don't see where it came from either.

shamil98 · Answer

Oh, never mind, the solution is the same thing!

shamil98 · Answer

$$2x(x-2)^4 + 4x^2(x-2)^3 = 2x(x-2)^3(3x-2)$$

shamil98 · Answer

I typed in the solution i got into wolfram, it's alternate form is the one from your book.

shamil98 · Answer

Although I'm still skeptical as to how it is the same thing lol.

anonymous · Answer

is there a way to prove those are equal somehow lol

shamil98 · Answer

We could solve it i guess.

anonymous · Answer

I guess I could use inductive reasoning and plug a 2 in there. that's a nono though

shamil98 · Answer

$$2x((x-2)^4 +2x(x-2)^3) = 2x((x-2)^3(3x-2)$$
dividing both sides by 2x
$$(x-2)^4 + 2x(x-2)^3 = (x-2)^3(3x-2)$$

shamil98 · Answer

From there i guess you would expand it and simplify.

anonymous · Answer

I plugged a 1 in for x and got 0=-2

shamil98 · Answer

I would expand it but it would be a huge huge equation..

shamil98 · Answer

$$(x-2)(x-2)(x-2)(x-2) + 2x(x-2)(x-2)(x-2)=(x-2)(x-2)(x-2)(3x-2)$$
and then you would multiply i guess.

anonymous · Answer

same here lol. if I plugged in 1 and it didn't work it cant be true. if you plug 0 in only . its true. does that mean anything to you?
couldn't we just start marking out x-2's on each side

shamil98 · Answer

We can't we have +2x(x-2)^3

shamil98 · Answer

we could have if it was multiplied by that ..

anonymous · Answer

$$2x(x-2)^4 + 4x^2(x-2)^3$$
$$=(x-2)^3(2x(x-2)+4x^2)$$

anonymous · Answer

both terms have a common factor of $(x-2)^3$

shamil98 · Answer

Oh wait, yes we can, i'm not thinking straight..

anonymous · Answer

the second factor is $4x^2-4x$ so maybe books answer is 
$$(x-2)^3(4x^2-4x)$$ or even 
$$4(x-2)^3(x^2-x)$$

shamil98 · Answer

$$(x-2)^3 ((x-2)+2x) = (x-2)^3(3x-2)$$
$$(x-2) + 2x = 3x-2$$
$$3x - 2 = 3x -2$$
so yes it is correct!

shamil98 · Answer

$$(x-2)^4 + 2x(x-2)^3 = (x-2)^3(3x-2)$$
As sattelite said the left side has a gcf of (x-2)^3
which is why it is
$$(x-2)^3 ((x-2)+2x)$$

anonymous · Answer

there must be many variations of this answer. Just need to have more faith in my answer. would you guys say that one of them is more 'correct' than the other?

shamil98 · Answer

I think the derivative that I got is the best, because you would have to simplify it a lot more to get the other way.

shamil98 · Answer

But, your book gave that simplified answer so next time i guess you are supposed to simplify it all the way

anonymous · Answer

possibly. thanks for the help. both of you. 3 different angles really helps

shamil98 · Answer

Anytime!