About derivatives of inverse functions

Question

Suppose $f(x)$ is the inverse function of $g(x)$.

Does that also imply that $\frac{df(x)}{dx}$ is also the inverse function of $\frac{dg(x)}{dx}$?

And can I thus use $g'(x)= \frac{1}{f'(g(x))}$ for calculating double derivatives too?

Answer 1

The inverse of $f(x) = e^x$ is $f^{-1}(x) = \ln x$, but the inverse of $f'(x) = e^x$ is not $\left(f^{-1}\right)'(x) = \frac 1x$. So no.

Answer 2

We start with $y = f(x)$. Differentiation with respect to $x$ gives $dy/dx = f'(x)$. One can also write this as $dy = f'(x) dx$.

Now take the inverse relation $x = g(y)$. Differentiation with respect to $y$ yields $dx/dy = g'(y)$, or alternatively $dx = g'(y)dy$.

Comparing these results we can conclude that $g'(y) = 1/f'(x)$. This is the correct relation between the derivative of a function and the derivative of the inverse of that function. You can easily verify that this works for the pair of functions $y = f(x) = e^x$ and $x = g(y) = \ln(y)$.

Answer 3

Suppose $\frac{\mathrm df}{\mathrm dx}$ is the inverse of $\frac{\mathrm dg}{\mathrm dx}$ for every $f(x)$ and its inverse $g(x)$, here is the result of it. For all $a\in\mathbf D_g$, there is (definition of inverse) $$\frac{\mathrm df}{\mathrm dx}\left(\frac{\mathrm dg}{\mathrm dx}(a)\right)=a$$

Since we all know $\frac{\mathrm df}{\mathrm dx}(g(a))\frac{\mathrm dg}{\mathrm dx}(a)=1$, it would be found that $$\frac{\mathrm df}{\mathrm dx}\left(\frac1{\frac{\mathrm df}{\mathrm dx}(g(a))}\right)=a$$ Let $g(a)=x$ and each function $f$, as you can see, should satisfy the condition below $$\frac{\mathrm df}{\mathrm dx}\left(\frac1{\frac{\mathrm df}{\mathrm dx}(x)}\right)=f(x)$$ However, $f$ does not need to meet such a complex condition to own its inverse $g$. Hence, your supposition does not hold and that formula can not be used to calculate the second derivative as well.