Question

The function f and g are differentiable at x=10 and x=20 and f(g(x))=x^2. if f(10)=5, f'(10)=4, f'(20)=-5, and g(10)=20, what is the value of g'(10)?

Answer 1

have you used chain rule to differentiate f(g(x)) yet?

Answer 2

I learned how to do the chain rule

Answer 3

ok then what is \[\frac{d}{dx}f(g(x))=?\]

Answer 4

f(x)g'(x)+g(x)f'(x)

Answer 5

that is the product rule
we don't have f*g
we have f composed with g

Answer 6

this is why I asked you to use chain rule to differentiate f(g(x))

Answer 7

I thought u use the chain rule when its something to a power

Answer 8

power rule is what you use when you have a constant power

Answer 9

chain rule is what you use when you have a function inside a function

Answer 10

ooh ok. So how would I do the chain rule for this?
would it be f'(g(x))*g'(x)?

Answer 11

that is right

Answer 12

\[f(g(x))=x^2 \\ \text{ differentiating both sides } \\ f'(g(x)) \cdot g'(x)=2x\]

Answer 13

not enter in 10 for x

Answer 14

\[f'(g(10)) \cdot g'(10)=2(10)\]

Answer 15

you are given g(10)

Answer 16

g(10)=20 right?

Answer 17

so replace g(10) with 20
\[f'(g(10)) \cdot g'(10)=2(10)\]
\[f'(20) \cdot g'(10)=2(10)\]

Answer 18

see if you can finish the rest

Answer 19

g'(10)=25?

Answer 20

\[f'(20)=-5 \\ \text{ so we have } \\ -5 \cdot g'(10)=2(10)\]

Answer 21

hmmm how did you get g'(10)=25?

Answer 22

oh I added, oops

Answer 23

you do know the operation between -5 and g'(10) is multiplication and not addition :p

Answer 24

g'(10)=-4?

Answer 25

20/-5 is -4
good worrk

Answer 26

thank you!

Answer 27

I have another question, do u mind helping me till?

Answer 28

I can take a look

Answer 29

In the table below, the values of f(x), g(x), f'(x) and g'(x) are given 2 values of x.
if y =[f(2x)+g(x)]^2, find y'(3)

Answer 30

well we know we are going to have differentiate since we want to find y'(of something)