The function $f(x)=\frac{2x^3+7x}{x^2+4}$ is invertible. Calculate the derivative $(f^{-1})(\frac{-9}{5})$. I was supposed to find the derivative of the inverse of the function f, and then its derivative and evaluate it at $\frac{-9}{5}$.
I initially tried by hand and it became really messy but I got a solution that seemed reasonable, however it was not correct. So I turned to WolframAlpha and got the same solution: $\frac{32761}{65047}$. I used the the following theorem: $(Df^{-1})(y)=\frac{1}{f'(x)}$.
Any suggestion on how to solve it, and/or perhaps if the theorem I'm using is useless in this case?
We have an online module that checks if your answer is correct so I don't know the actual solution.
Thanks :)

If $f^{-1}$ exists, then
$y=f(x), y_0=f(x_0) \implies x_0=f(y_9)$ Here for $y_0=-9/5$ $x_0=-1 \implies f^{-1}(-9/5)=-1.$
$$f(x)=\frac{2x^3+7x}{x^2+4} \implies f'(x)=2-\frac{8}{x^2+4}^2+\frac{1}{x^2+4}$$
$$\frac{d f^{-1}(y)}{dy}|_{y=y_0}=\frac{1}{f'(x_0)} \implies \frac{d f^{-1}(y)}{dy}|_{y=-9/5}=\frac{25}{47}.$$
