Question

How am I supposed to do this!!??
F=f o g (f times g)
H(x)=x(x-2f(x))^4
g(3)=-2
g'(3)=4
f(3)=2
f'(3)=1
f'(-2)=-1
Find F'(3) and H'(3)???

Answer 1

The product rule of differentiation. What is the derivative of fg, just in general terms?

Answer 2

f(x) times g'(x) + g(x) times f'(x)?

Answer 3

Yes, so use that for F = fg

Answer 4

ok hold on lemme see if i get it

Answer 5

i got the wrong answer...what am i doing wrong? i did 2(4)+-2(1)?

Answer 6

well the way i read your problem, it is not f times g
it looks like
\[F=f\circ g\] not
\[F=f\times g\]

Answer 7

yea its the first one with the circle

Answer 8

so you need the "chain rule" rather then the product rule

Answer 9

In which case you need the chain rule

Answer 10

i forget the chain rule...

Answer 11

\[(f\circ g)'=f'(g)\times g'\]

Answer 12

or if you prefer
\[(f\circ g)'(x)=f'(g(x))\times g'(x)\] try those numbers for
\[(f\circ g)'(2)=f'(g(2))\times g'(2)\]

Answer 13

shouldnt i be doing it for 3not 2?

Answer 14

oh sorry it is
\[(f\circ g)'(3)=f'(g(3))\times g'(3)\]

Answer 15

yes my mistake

Answer 16

haha okay so how would i do g(3)=-2 inside of f'(3)=1? 1 times -2?

Answer 17

(I should have seen that this was the composite, thanks sat73)

Answer 18

how do i do f'(g(3))?

Answer 19

Well, g(3) = -2, hence f'(g(3)) = f'(-2) = ...

Answer 20

ohh okayy