This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We prove this from the definition of the derivative in terms of limits.

Let\ y=f(x)\\\\ x=f^{-1}(y)\\\\
Let us say x increases by Δx amount and consequently,  y increases by Δy, ie., y becomes (y+Δy).   Thus derivative :

\frac{dy}{dx}= \lim_{\Delta x \to 0} \frac{(y+\Delta y)-y}{(x+\Delta x)-x} =\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}\\

The converse is also true, about the changes in y and x. When y changes from y to (y+Δy) then due to the same relationship between  x and y,  x will change from  x to (x+Δx).  So derivative:

 \frac{dx}{dy}= \lim_{\Delta y \to 0} \frac{(x+\Delta x)-x}{(y+\Delta y)-y} =\lim_{\Delta y \to 0} \frac{\Delta x}{\Delta y}\\\\

Hence the product:
\frac{dy}{dx}*\frac{dx}{dy}=\lim_{\Delta y \to 0} \frac{\Delta y}{\Delta x}*\lim_{\Delta y \to 0} \frac{\Delta x}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta x \to 0} \frac{\Delta y}{\Delta x}* \frac{\Delta x}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta y \to 0} \frac{\Delta x}{\Delta x}* \frac{\Delta y}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta y \to 0}1*1 =1,\ \ As, however\ small\ \Delta x\ is\ \frac{\Delta x}{\Delta x}\ is\ 1\\\\ Hence, \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\\ 
Alternately,  the same can be proved using the y = f(x) and y+Δy = f(x+Δx) using the definition of derivatives by  limits.