In this post, I derive the so-called product rule that is taught in a Calculus course. The product rule enables one to find the derivative of a function which can be expressed as a product of two functions. That is,  the product rule allows for evaluating

$\frac{d h(x)}{dx} \big|_{a^+}$

with $h(x) \equiv f(x) g(x)$.

To start, use the corresponding definition of a derivative from this post :

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{h(a + \Delta x) – h(a)}{\Delta x}$

Substitute $h(x) \equiv f(x) g(x)$ :

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{f(a + \Delta x)g(a + \Delta x) – f(a)g(a)}{\Delta x}$

Now I approximately follow the outline in Reference 1. The next step is to add zero to the numerator of the previous equation, as follows.

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{f(a + \Delta x)g(a + \Delta x) – f(a)g(a) + f(a + \Delta x)g(a) – f(a + \Delta x)g(a)}{\Delta x}$

Now factor.

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{f(a + \Delta x) \bigg( g(a + \Delta x) – g(a) \bigg) + g(a) \bigg( f(a + \Delta x) – f(a) \bigg) }{\Delta x}$

Rewrite.

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \bigg[ \frac{f(a + \Delta x) \bigg( g(a + \Delta x) – g(a) \bigg)}{\Delta x} + \frac{g(a) \bigg( f(a + \Delta x) – f(a) \bigg) }{\Delta x} \bigg]$

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{f(a + \Delta x) \bigg( g(a + \Delta x) – g(a) \bigg)}{\Delta x} + \lim_{\Delta x \rightarrow 0^+} \frac{g(a) \bigg( f(a + \Delta x) – f(a) \bigg) }{\Delta x}$

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} f(a + \Delta x) \lim_{\Delta x \rightarrow 0^+} \frac{ g(a + \Delta x) – g(a) }{\Delta x} + \lim_{\Delta x \rightarrow 0^+} g(a) \lim_{\Delta x \rightarrow 0^+} \frac{ f(a + \Delta x) – f(a) }{\Delta x}$

Use the definition of a derivative written in this post.

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} f(a + \Delta x) \frac{dg(x)}{dx}\big|_{a^+} + \lim_{\Delta x \rightarrow 0^+} g(a) \frac{df(x)}{dx}\big|_{a^+}$

Using this property, $\lim_{\Delta x \rightarrow 0^+} g(a) = g(a)$, so

$\frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} f(a + \Delta x) \frac{dg(x)}{dx}\big|_{a^+} + g(a) \frac{df(x)}{dx}\big|_{a^+}$.

Assume that $\lim_{\Delta x \rightarrow 0^+} f(a + \Delta x) = f(a)$. Then

$\frac{d h(x)}{dx} \big|_{a^+} = f(a) \frac{dg(x)}{dx}\big|_{a^+} + g(a) \frac{df(x)}{dx}\big|_{a^+}$.

Finally, set $a=x$. Then

$\boxed{ \frac{d h(x)}{dx} \big|_{x^+} = f(x) \frac{dg(x)}{dx}\big|_{x^+} + g(x) \frac{df(x)}{dx}\big|_{x^+} }$.

References:

[1] https://tutorial.math.lamar.edu/classes/calci/DerivativeProofs.aspx