Calc II Question: Inverse Functions and Derivatives Explained | Help Needed

[frasifrasi]

This deals with inverse functions:

suppose g(x) is the inverse of f(X) and G(X) = 1/g(X). If f(3) = 3 and f'(3) = 1/9, find G'(3).

Does anyone know how to answer this question?

Thanks.

I was thinking of using the formula g'(x) = 1/f'(g(X)), but the G(X) is throwing me off.

 

[ptr]

$$f'(x)= \frac{1}{f^{-1}'(x)}$$

 

[frasifrasi]

tthat is not what is being asked...

 

[HallsofIvy]

This deals with inverse functions:

suppose g(x) is the inverse of f(X) and G(X) = 1/g(X). If f(3) = 3 and f'(3) = 1/9, find G'(3).

Does anyone know how to answer this question?

Thanks.

I was thinking of using the formula g'(x) = 1/f'(g(X)), but the G(X) is throwing me off.

If G(x)= 1/ g(x)= g^-1 then, by the chain rule, G'= -1g(x)^-2 g'(x). Since g(x) is f^-1(x), g'(x)= 1/f'(x).

 

[frasifrasi]

I still don't get it. Can anyone explain it using the actual numbers to derive an answer? I need to understand this before the exam.