What Is g'(2) for the Function G = (1/f^-1)?

[Geologist180]

Homework Statement

1. Suppose that f has an inverse and f(-4)=2, f '(-4)=2/5. If G= (1/f^-1) what is g '(2) ?

If it helps the answer is (-5/32)

Homework Equations

[/B]
f^-1'(b)=1/(f')(a)

The Attempt at a Solution

Im not really sure how to start this problem. I am familiar with how to use the equation above to plug solve for "c" and then plug c into the derivative, so I assume it has something to do with that but I can't find any helpful resources on the web showing me how to do this specific type of problem.[/B]

 

[LCKurtz]

Homework Statement

1. Suppose that f has an inverse and f(-4)=2, f '(-4)=2/5. If G= (1/f^-1) what is g '(2) ?

I suppose you mean G'(2)

If it helps the answer is (-5/32)

Homework Equations

[/B]
f^-1'(b)=1/(f')(a)

The Attempt at a Solution

Im not really sure how to start this problem. I am familiar with how to use the equation above to plug solve for "c" and then plug c into the derivative, so I assume it has something to do with that but I can't find any helpful resources on the web showing me how to do this specific type of problem.[/B]

Start by showing us the calculation for G' using the chain rule.

 

[Geologist180]

I suppose you mean G'(2)
Start by showing us the calculation for G' using the chain rule.

I believe the derivative of G would be -1/(f^-1) ²
Is this correct?

 

[LCKurtz]

No. You haven't used the chain rule on the definition of G.

 

[Geologist180]

Im confused how I am supposed to use the chain rule in the context of this question.
When I think about the chain rule, I think about more complicated functions and taking the derivative of the functions with respect to the outside of the brackets multiplied by the derivative of the inside of the brackets. I am having trouble seeing how to apply that here.

 

[LCKurtz]

Im confused how I am supposed to use the chain rule in the context of this question.
When I think about the chain rule, I think about more complicated functions and taking the derivative of the functions with respect to the outside of the brackets multiplied by the derivative of the inside of the brackets. I am having trouble seeing how to apply that here.

Isn't that exactly what you have:
$$
G(\cdot)=(f^{-1}(\cdot))^{-1}$$where you can replace the dot with whatever variable you want to call it, maybe $y$ if you call the original equation $y=f(x)$. What is the derivative of the "inside"?

I know it is a bit confusing with one of the $-1$'s being a negative exponent and the other an inverse function.

 

[Geologist180]

Isn't that exactly what you have:
$$
G(\cdot)=(f^{-1}(\cdot))^{-1}$$where you can replace the dot with whatever variable you want to call it, maybe $y$ if you call the original equation $y=f(x)$. What is the derivative of the "inside"?

I know it is a bit confusing with one of the $-1$'s being a negative exponent and the other an inverse function.

I believe it would be: (substituted a star for the dot)

( - (f^-1(*))^-2) (d/dx (f^-1(*)))

Am I following you correctly in this?
Thank you for your hep by the way

 

[LCKurtz]

I believe it would be: (substituted a star for the dot)

( - (f^-1(*))^-2) (d/dx (f^-1(*)))

Am I following you correctly in this?
Thank you for your hep by the way

With the * that would be written ( - (f^-1(*))^-2) (d/d* (f^-1(*))). Let's use $y$ and write it$$
G'(y)=-(f^{-1}(y))^{-2}\frac d {dy} f^{-1}(y)$$Now, remember your question is asking for $G'(2)$. If you put that in you have$$
G'(2) =-(f^{-1}(2))^{-2}\frac d {dy} f^{-1}(2)$$So now you have to ask yourself: do I know $f^{-1}(2)$? Do I know the derivative of the inverse function at $2$? What do I know about derivatives of inverses versus derivatives of the original function?