Derivative of the Inverse function

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

$x=g\left(y\right)=g\left(f\left(x\right)\right)$  and $y=f\left(x\right)=f\left(g\left(y\right)\right)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Consider the function $f\left(x\right)={x}^{3}+4x$, which has an inverse function ${f}^{-1}\left(x\right)$. 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 $\left({f}^{-1}{\right)}^{\prime }\left(5\right)$.

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

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. :-)