Question

if f(x)=(x+1)^2(x+2)^3 find f '(0)

Answer 1

Find the derivative first then plug in 0. Or do you have to do it with the difference quotient?

Answer 2

find the derivative

Answer 3

We do you know how?

Answer 4

*Well

Answer 5

will that be 2(x+1)^3

Answer 6

3(x+2)^4

Answer 7

Actually, I think this might be product rule with chain rule..

Answer 8

oh ok , so how can i find the derivative when (x+1)^2 is just just 1

Answer 9

Okay, so we have:
\[\LARGE f(x)=(x+1)^2(x+2)^3\]
We have to use a combo of product rule with chain rule, but lucky for us it would just be 1..
\[\LARGE f'(x)=[(x+1)^2*3(x+2)^2*1]+[(x+2)^3*2(x+1)*1]\]

Do understand what I did?

Answer 10

*
\[\large f'(x)=[(x+1)^2*3(x+2)^2*1]+[(x+2)^3*2(x+1)*1]\]

Answer 11

yes

Answer 12

Okay, now just simplify that big mess and you're done :)
@zepdrix Just checking, did I do this right? >.<

Answer 13

ok solving now

Answer 14

i pluged in zero for my x and i got 47 which is not a listed answer

Answer 15

Btw, you don't have to like distribute everything out, unless your instructor says so(in that case they'd be the cruelest >:( )
But you could just leave it as:
\[\large f'(x)=3[(x+1)^2(x+2)^2]+2[(x+2)^3(x+1)]\]

Answer 16

ok i got the answer 24

Answer 17

thats listed

Answer 18

Plug in 0..
\[\LARGE 3((1)^2(2)^2)+2(2)^3(1))\]
\[\LARGE 12+16\]
\[\LARGE 28\]
:l

Answer 19

i made a mistake with the 2^2

Answer 20

i put 8 calculation error

Answer 21

But , yea, it's 28. Hope that made sense to ya >.<

Answer 22

yes it did i would got the same answer if i didn't calculate wrong