Inverse Derivative: Error/Intuition

Question

Problem is to find $[f^{-1}(4)]'$ given $f(x)=\frac{x^3+7}{2}$.

Way One: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}$. Therefore, $\sqrt[3]{2x-7}=y$ (i.e. our inverse function). So, $f^{-1}(x)=(2x-7)^{\frac{1}{3}}$. So, $\frac{d}{dx}f^{-1}(x)=\frac{2}{3}(2x-7)^{-\frac{2}{3}}\implies \frac{d}{dx}f^{-1}(4)=\frac{2}{3}.$

Way Two: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}\implies 2x=y^3+7\implies 2=3y^2\frac{dy}{dx}\implies \frac{1}{\frac{3}{2}y^2}=\frac{dy}{dx}$.

Now, this to me seems like the inverse function is $\frac{d}{dx}f^{-1}|_{x=3}=\frac{dy}{dx}|_{x=3}=\frac{1}{\frac{3}{2}y^2}$ when $x=3$ we know $y=f(3)$ which isn't our answer. So, what am I doing wrong here? Can someone explain what is incorrect with this reasoning? I feel like I am missing something when doing these problems.

Answer 1

It was tricky for me too so I started doing these problems the following way. You know that $f(f^{-1}(x)) = x$ so differentiating both sides gives you $$f'(f^{-1}(x)) \cdot \frac{d}{dx} f^{-1}(x) = 1 \implies \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x)) }.$$ Now $f^{-1}(4) = 1$ so $$\frac{d}{dx} f^{-1}(4) = \frac{1}{f'(f^{-1}(4)) } = \frac{1}{f'(1)} = \frac{1}{3(1)^2/2} = \frac{2}{3}.$$

Answer 2

You already know that after switching variables \begin{align*} \frac{dy}{dx}=\frac{1}{\frac{3}{2}y^2}\tag{1} \end{align*} Since we want to know $\left(f^{-1}(4)\right)^{\prime}$ we obtain again after switching variables from $x=\frac{y^3+7}{2}$ evaluated at $x=4$:

\begin{align*} 4&=\frac{y^3+7}{2}\\ 8&=y^3+7\\ y&=1 \end{align*} putting it into (1) we get \begin{align*} \left.\frac{dy}{dx}\right|_{y=1}&=\left.\frac{1}{\frac{3}{2}y^2}\right|_{y=1} \color{blue}{=\frac{2}{3}} \end{align*} in accordance with the first way.

Note: It might be easier to calculate way two without switching variables after all.