> [!summary] This proof uses the definition of a derivative to prove the difference 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 difference 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} $ --- > ✍️ This project’s been a labour of love. > If it helped, [give Math & Matter a star](https://github.com/rajeevphysics/Obsidian-MathMatter) and let me know what you'd like to see next. ---