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)?

- amy0799

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- freckles

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

- amy0799

I learned how to do the chain rule

- freckles

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

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## More answers

- amy0799

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

- freckles

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

- freckles

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

- amy0799

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

- freckles

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

- freckles

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

- amy0799

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

- freckles

that is right

- freckles

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

- freckles

not enter in 10 for x

- freckles

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

- freckles

you are given g(10)

- freckles

g(10)=20 right?

- freckles

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

- freckles

see if you can finish the rest

- amy0799

g'(10)=25?

- freckles

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

- freckles

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

- amy0799

oh I added, oops

- freckles

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

- amy0799

g'(10)=-4?

- freckles

20/-5 is -4
good worrk

- amy0799

thank you!

- amy0799

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

- freckles

I can take a look

- amy0799

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)

- freckles

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

- amy0799

|dw:1441749521965:dw|

- freckles

again here you will have to use chain rule
this chain rule does come with power rule because we have have a constant power
we will also have to use sum rule
then chain rule again

- amy0799

2(f(2x)+g(x))*(f'(2x)+g'(x))

- freckles

ok I only have one complaint...

- freckles

let's look at f(2x)

- freckles

notice the inside function is 2x

- freckles

the outside function is f( )

- freckles

when you find the derivative of f(2x) w.r.t. x you should get 2*f'(2x)
derivative of inside time derivative of outside

- freckles

\[y(x)=[f(2x)+g(x)]^2 \\ y'(x)=2[f(2x)+g(x)] \cdot [2 f'(2x)+g'(x)]\]

- freckles

now we want to find y'(3)

- freckles

so replace x with 3

- freckles

\[\\ y'(3)=2[f(2\cdot 3)+g(3)] \cdot [2 f'(2 \cdot 3)+g'(3)]\]

- amy0799

y'(3)=112

- freckles

\[f(6)=? \\ g(3)=? \\ f'(6)=? \\ g'(3)=?\]
you found these values from the table ?

- freckles

\[f(6)=-3 \\ g(3)=-1 \\ f'(6)=-2 \\ g'(3)=-5\]
this is what I see for those values
is that what you also see

- amy0799

oh I know what I did wrong. Hold on

- amy0799

y'(3)=72?

- freckles

purrrfect
(you know like a cat)

- amy0799

haha thank you so much for the help!

- freckles

np
I hope everything makes more sense

- amy0799

Yup it does thanks to your help :D

- freckles

np
I must go now
good luck with calculus

- amy0799

ok. thanks, I need all the luck I can get haha

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