> [!summary]
This proof uses definitions from derivatives to prove the sum rule.
>
**Key Result:**
$\frac{d}{dx}[f (x)+g (x)] = \frac{d}{dx}f (x)+\frac{d}{dx}g (x)$
>[!info]+ Read Time
**⏱ 1 min**
# Mathematical Proof
> [!warning] Assumptions
To prove the sum rule using [[Direct Proof|direct proof]] assume the following:
> - The definition of a [[Derivative|derivative]] is $\frac{dy}{dx} = f'(x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h}$
Prove that $\frac{d}{dx}[f (x)+g (x)] = \frac{d}{dx}f (x)+\frac{d}{dx}g (x)$.
$
\begin{array}{c}
\text{From the definition of a derivative:} \\ \\
\begin{align*}
\frac{d}{dx}[f(x)+g(x)] &=\displaystyle \lim_{ h \to 0 } \frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h} \\
&= \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h} \\
&= \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h}+ \displaystyle \lim_{ h \to 0 } \frac{g(x+h)-g(x)}{h}
\end{align*} \\ \\
\text{Then by definition $\frac{d}{dx}[f (x)+g (x)] = \frac{d}{dx}f (x)+\frac{d}{dx}g (x)$}
\end{array}
$
# Resources
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