Let $y=f(x)$ and $x=g(y)$ are inverse functions of each other. Then

$x=g(y)=g(f(x))$ and $y=f(x)=f(g(y))$

Considering $x=g(y)=g(f(x))$ and differentiating both sides with respect to $x$ , we get

$1={g}^{\prime}(f(x))\cdot {f}^{\prime}(x)$

or, $1={g}^{\prime}(y)\cdot {f}^{\prime}(x)$

Similarly, considering $y=f(x)=f(g(y))$ and differentiating both sides with respect to $y$ , we get

$1={f}^{\prime}(g(y))\cdot {g}^{\prime}(y)$

or, $1={f}^{\prime}(x)\cdot {g}^{\prime}(y)$

So, it does not matter whether we consider $g(f(x))$ or, $f(g(y))$, we arrive at the same relation between the derivatives of the original function $y=f(x)$ and the inverse function $g(y)$

$1={f}^{\prime}(x)\cdot {g}^{\prime}(y)$

or, ${g}^{\prime}(y)=\frac{1}{{f}^{\prime}(x)}$

If we write $\frac{dx}{dy}$ for ${g}^{\prime}(y)$ and $\frac{dy}{dx}$ for ${f}^{\prime}(x)$ then we have

$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$

The usual notation for the inverse of the function $y=f(x)$ is given by $g(y)={f}^{-1}(y)$

So, what we have is $({f}^{-1}{)}^{\prime}(y)=\frac{1}{{f}^{\prime}(x)}$

What this tells us is that if we know the value of the derivative of the function $f(x)$ at $x=a$ , then we can easily calculate the derivative of its inverse function ${f}^{-1}(y)$ at the point $y=b=f(a)$ as follows:

$({f}^{-1}{)}^{\prime}(b)=\frac{1}{{f}^{\prime}(a)}$

Consider the function $f(x)={x}^{3}+4x$, which has an inverse function ${f}^{-1}(x)$. It is not easy to find the exact definition of this inverse function. Still, without knowing the exact definition of this inverse function, we can determine the value of $({f}^{-1}{)}^{\prime}(5)$.

Since $f(1)=5$, we have $({f}^{-1}{)}^{\prime}(5)=\frac{1}{{f}^{\prime}(1)}=\frac{1}{7}$. We have used the fact that ${f}^{\prime}(x)=3{x}^{2}+4$ and ${f}^{\prime}(1)=7$.

While we have to be careful in what we say, but we can think of it this way. The derivative of the inverse is the inverse of the derivative. :-)

Also, think of it this way. Suppose A is moving 4 times faster than B. You can construct a function to describe this. Now think of the function that describes how fast is B moving with respect to A. Clearly, this function is the inverse of the original function. And, you can say B is moving $\frac{1}{4}$ times faster than A. :-)