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The problem says:

If $f(x)=\frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.

I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ but I dont know how.

The answer says (f^-1)'(2)=1/4

Thanks

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Hint: Since $f(1)=2$, $f^{-1}(2)=1$.

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  • $\begingroup$ ok, but how did you solve 2=f(x) for x? $\endgroup$ – L. G. Dec 15 '18 at 12:47
  • $\begingroup$ $$\frac{4x^3}{x^2+1}=2\iff4x^3=2x^2+2.$$Now, use the rational root theorem. $\endgroup$ – José Carlos Santos Dec 15 '18 at 12:49
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If $y=f(x)$ we also have $dy/dx=f'(x)$. If $f$ is one to one, we have $x=g(y)$ for some $g$.

We also have that $dy/dx=f'(x)$, $dx/dy=g'(y)$, where prime indicates taking a derivative with respect to the respective independent variable.

The derivatives are reciprocals of each other.

So we can find the derivative of the inverse at 2 by first finding what x gives us y=2, finding the derivative with respect to x at that value of x, then taking the reciprocal.

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