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Mathematics 16 Online
OpenStudy (anonymous):

if g(f(x))=x, g(5)=2 and g'(5)=14, then f'(2) is ?? help, please!

myininaya (myininaya):

Do you know chain rule

OpenStudy (anonymous):

yes, isnt it nu'u^n-1

myininaya (myininaya):

(g(f(x)))'=g'(f(x))f'(x)

myininaya (myininaya):

So if we differentiated one side you must do the same to the side

myininaya (myininaya):

Other side*

myininaya (myininaya):

What is f(2)?

OpenStudy (anonymous):

wait...how would i plug that in? so would it be g(f(x))'=14(x)(2)??

myininaya (myininaya):

Ok no . Do you know how to find f(2)?

OpenStudy (anonymous):

no

myininaya (myininaya):

So we have g(f(x))=x

myininaya (myininaya):

Plug in x=2

myininaya (myininaya):

So we have g(f(2))=2. Assuming g has an inverse we could say f(2)=g^(-1)(2). But we know g(5)=2 so g^(-1)(2)=5. So we have f(2)=5 right?

myininaya (myininaya):

Now back to the derivative part we had g'(f(x))f'(x)=1

myininaya (myininaya):

Now plug in 2 for x

myininaya (myininaya):

So you would have g'(f(2))f'(2)=1

myininaya (myininaya):

War is f(2) again ?.

myininaya (myininaya):

What*

myininaya (myininaya):

We almost there

myininaya (myininaya):

All we have to do is plug in

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