In this post, the chain rule is proved. This rule frequently appears in Calculus. Recall from this post that: $dx|_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \Delta x$ and $df(x)|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \Delta f (\Delta x)$. Suppose a variable $y$ can be written as a function of another variable $u$, and that $u$ can be written as a function of another variable $x$. Then $y$ can be written as a function of $x$: $y(x)$. What is $\frac{dy(x)}{dx}\bigg|_{a^+}$ if only $y(u)$ and $u(x)$ are known? Solution: Start by writing $\frac{dy(x)}{dx}\bigg|_{a^+} = \frac{dy(x)}{dx}\bigg|_{a^+}$. According to this post, the $|_{a^+}$ symbols can be moved as follows. $\frac{dy(x)}{dx}\bigg|_{a^+} = \frac{dy(x) |_{a^+} }{dx |_{a^+} }$. Using the information at the top of the current post, $\frac{dy(x)}{dx}\bigg|_{a^+} = \frac{ \lim_{\Delta x \rightarrow 0^+} \Delta y (\Delta x) }{ \lim_{\Delta x \rightarrow 0^+} \Delta x}$. Similarly, $ du(x) \big|_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \Delta u(\Delta x)$. From this post, $ \frac{dr}{dr} \bigg|_{r^+} = 1$. Replacing $r$ by the arbitrary function $u(x)$ and replacing $r^+$ by $a^+$ yields $ \frac{du(x)}{du(x)}\bigg|_{a^+} = 1 = \frac{du(x) |_{a^+}}{du(x) |_{a^+} }$. Recall: $\frac{dy(x)}{dx} \bigg|_{a^+} = \frac{dy(x) |_{a^+} }{ dx|_{a^+} }$. Multiply the right side by 1: $ \frac{dy(x)}{dx}\bigg|_{a^+} = \frac{dy(x)|_{a^+} }{dx |_{a^+}} \frac{du(x) |_{a^+}}{du(x) |_{a^+}}$. From the commutative property for multiplication, the previous equation becomes: $ \frac{dy(x)}{dx} \bigg|_{a^+} = \frac{dy(x)|_{a^+} }{ du(x) |_{a^+}} \frac{du(x) |_{a^+}}{dx |_{a^+}} $. After moving the $|_{a^+}$ symbols, the previous equation becomes: $ \boxed{ \frac{dy(x)}{dx} \bigg|_{a^+} = \frac{dy(x) }{ du(x) }\bigg|_{a^+} \frac{du(x) }{dx} \bigg|_{a^+}  }$. This is the chain rule. $\Box$