Question

Graphs and derivatives!

Answer 1

help!

Answer 2

@dan815

Answer 3

are you able to determine what `P ' (x)` is?

Answer 4

Em, you mean with the product rule?

Answer 5

yes

Answer 6

p'(x) = f'(x)g(x)+f(x)g'(x)

Answer 7

yep

Answer 8

now replace every x with 2

Answer 9

P ' (x) = F '(x)*G(x) + F(x)*G ' (x)
P ' (2) = F '(2)*G(2) + F(2)*G ' (2)

Answer 10

F(2) and G(2) are simply the y coordinates on F(x) and G(x) respectively when x = 2

Answer 11

F ' (2) is the slope of the tangent line on F(x) when x = 2
G ' (2) is the slope of the tangent line on G(x) when x = 2

Answer 12

Wait wait, so now we have
p'(2)=f'(2)g(2)+f(2)g'(2)

Answer 13

f(2) = 3
g(2)=2

Answer 14

`f(2) = 3`
`g(2)=2`
yes on both

Answer 15

f'(2) = 0
g'(2) = ..?

Answer 16

ooh oh
g'(2) = 1/2

Answer 17

yes on those two as well

Answer 18

p'(2) = 3/2 ?

Answer 19

yep

Answer 20

q'(x) = 3/8?

Answer 21

hrrm no I did it wrong.

Answer 22

Where did I do it wrong?
q'(x) = [g(x)f'(x) - f(x)g'(x)]/g(x)^2

Answer 23

q'(x) = [2(0)-3(1/2)]/4

Answer 24

we want Q ' (7), not Q ' (2)

Answer 25

oou...

Answer 26

q'(7) = 43/12?

Answer 27

yep

Answer 28

Yay!! thank you :)

Answer 29

you're welcome