> [!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 <iframe width="560" height="315" src="https://www.youtube.com/embed/sca0v-gVcMo?si=2mFKFgnaqZq3EY6O" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> --- > 📚 Like this note? [Star the GitHub repo](https://github.com/rajeevphysics/Obsidian-MathMatter) to support the project and help others discover it! ---