Answer
a) If
$$y=f^{-1}(x),$$
then
$$f(y)=x .$$
Differentiating implicitly with respect to $x$ and remembering that $y$ is a function of $x$ we get:
$$ f^{\prime}(y) \frac{d y}{d x}=1,$$
so
$$\frac{d y}{d x}=\frac{1}{f^{\prime}(y)} $$
$\Rightarrow$
$$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$$.
b)
$$ f(4)=5 \quad \Rightarrow \quad f^{-1}(5)=4 .$$
By part }(a),
$$
\left(f^{-1}\right)^{\prime}(5)=\frac{1}{f^{\prime}\left(f^{-1}(5)\right)}=\frac{1}{f^{\prime}(4)}=1 /\left(\frac{2}{3}\right)=\frac{3}{2}$$
Work Step by Step
a) If
$$y=f^{-1}(x),$$
then
$$f(y)=x .$$
Differentiating implicitly with respect to $x$ and remembering that $y$ is a function of $x$ we get:
$$ f^{\prime}(y) \frac{d y}{d x}=1,$$
so
$$\frac{d y}{d x}=\frac{1}{f^{\prime}(y)} $$
$\Rightarrow$
$$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$$.
b)
$$ f(4)=5 \quad \Rightarrow \quad f^{-1}(5)=4 .$$
By part (a),
$$
\left(f^{-1}\right)^{\prime}(5)=\frac{1}{f^{\prime}\left(f^{-1}(5)\right)}=\frac{1}{f^{\prime}(4)}=1 /\left(\frac{2}{3}\right)=\frac{3}{2}$$
