Derivative of an inverse question

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SUMMARY

The discussion focuses on finding the derivative of the inverse function h(x) for f(x) = x³ + x at the point where h(2) = 1. The key equation used is (F^{-1})'(x) = 1 / F'(F^{-1}(x)). The derivative f'(x) is calculated as f'(x) = 3x² + 1, and since f(1) = 2, it follows that h(2) = 1. Therefore, h'(2) can be computed as h'(2) = 1 / f'(1), leading to the conclusion that h'(2) = 1 / (3(1)² + 1) = 1/4.

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Homework Statement



f(x)=(x^3)+x. If h(x) is the inverse of f(x), find h'(2).


Homework Equations



(F[tex]^{-1}[/tex])'(x)=[tex]\frac{1}{F'(F^{-1}x)}[/tex]

The Attempt at a Solution




I want to find h'(2)=(F[tex]^{-1}[/tex])'(2)=[tex]\frac{1}{F'(F^{-1}(2))}[/tex]

I know f'(x)=3(x^2)+1, so I just need to find h(2), but I don't know how to solve f(x)=(x^3)+x for its inverse. Is it possible to solve for x and then switch x and y with this type of function?
 
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It might be difficult finding the inverse function, but it's easy to see that f(1)=2, so h(2)=1.
 
Since h is the inverse of f, h(f(x)) = x
Differentiating, you get h'(f(x))*f'(x) = 1, so h'(f(x)) = 1/f'(x).

Can you work in grief's comment that f(1) = 2 to find h'(2)? You will also need to find f'(x), and from that f'(1).
 

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