Derivative of an inverse question

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

The Attempt at a Solution




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

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).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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