> [!summary]
This proof uses the definition of a derivative to prove the product rule.
>
Key Result:
$\frac{d}{dx}[f(x)\cdot g(x)] = f'(x)g(x)+f(x)g'(x)$
>[!info]+ Read Time
**⏱ 2 mins**
# Mathematical Proof
> [!warning] Assumptions
To prove the derivative product rule through [[Direct Proof|direct proof]], assume the following:
> - The definition of a [[Derivative|derivative]] is $\frac{dy}{dx} = f'(x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h}$
> - The [[Analytical Limits Rules|sum of two limits]] is $\displaystyle \lim_{ x \to a } [f (x)+g (x)]=L+M$
> - The product of two limits is $\displaystyle \lim_{ x \to a } [f(x)\cdot g(x)]=L\cdot M$
Prove $\frac{d}{dx}[f(x)\cdot g(x)] = f'(x)g(x)+f(x)g'(x)$
$
\begin{array}{c}
\frac{d}{dx}[f(x)\cdot g(x)] = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x+h)-f(x)g(x)}{h} \\ \\ \\
\text{Add and subtract the limit by $f(x+h)g(x)$ so the limit is now:}\\
= \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h} \\
\\
\text{Split into two limits}\\
= \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x+h)-f(x+h)g(x)}{h} + \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x)-f(x)g(x)}{h} \\
= \displaystyle \lim_{ h \to 0 } f(x+h)\frac{g(x+h)-g(x)}{h} + \displaystyle \lim_{ h \to 0 } g(x)\frac{f(x+h)-f(x)}{h} \\ \\
\text{Spread limit and simplify to get the following:}\\
= f(x)\displaystyle \lim_{ h \to 0 } \frac{g(x+h)-g(x)}{h} + \displaystyle g(x)\lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h} \\ \\
\text{Which is saying:} \\
f(x)g'(x)+ g(x)f'(x)
\end{array}
$
# Resources
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