Inverse of function and derivative

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

2 Answers

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

• ok, but how did you solve 2=f(x) for x? – L. G. Dec 15 '18 at 12:47
• $$\frac{4x^3}{x^2+1}=2\iff4x^3=2x^2+2.$$Now, use the rational root theorem. – José Carlos Santos Dec 15 '18 at 12:49

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.