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.

$\Box$