I saw proof of chain rule and I must say that I don't understand one key step.
Here is the proof: Consider function $f(x)$ and tangent $t.$ Let $\frac{dy}{dx}$ be slope of a tangent line in the point $A(a, f(a)).$ Consider another point, $B(a + \Delta{x}, f(a + \Delta{x}).$ Then slope of a secant line AB is $\frac{\Delta{y}}{\Delta{x}}.$ Here is of course $dx = \Delta{x}.$ We can now assume that $dy - \Delta{y} = \epsilon.$ From here is easy to see that $\Delta y = f'(a) \cdot \Delta x - \epsilon$
Finally let $y = f(g(x)), u = g(x)$ and $g(a) = b.$ Then $\Delta y = f'(b) \cdot \Delta u - \epsilon_1, \Delta u = g'(a) \cdot \Delta x - \epsilon_2.$
Combining this together we get $\Delta y = f'(b) \cdot (g'(a) \cdot \Delta x - \epsilon_2) - \epsilon_1.$
We can divide both sides by $\Delta x$: $\frac{\Delta y}{\Delta x} = f'(b) \cdot g'(a) - \frac{\epsilon_2}{\Delta x} \cdot f'(b) - \frac{\epsilon_1}{\Delta x}.$
Now, if $\Delta x \rightarrow 0,$ then also $\frac{\epsilon_1}{\Delta x}, \frac{\epsilon_2}{\Delta x} \rightarrow 0$ (*) I don't understand this part, why exactly we can conclude that?
Proof finished with ...From here using (*) we get $\frac{dy}{dx} = f'(b) \cdot g'(a).$