> [!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} $ --- <!-- Light Mode Newsletter Embed --> <div class="mm-form-light"> <iframe src="https://updates.cyberleadhub.com/widget/form/Y0kpQVpjJQuxEfX59m17" id="inline-Y0kpQVpjJQuxEfX59m17" title="Join Math & Matter Newsletter (Light)" data-height="900" scrolling="no" allowtransparency="true" loading="lazy" style="width:100%;height:350px;border:none;border-radius:10px;background:transparent;overflow:hidden" ></iframe> </div> <!-- Dark Mode Newsletter Embed --> <div class="mm-form-dark"> <iframe src="https://updates.cyberleadhub.com/widget/form/lbeDLm24VjuaFxhjccA1" id="inline-lbeDLm24VjuaFxhjccA1" title="Join Math & Matter Newsletter (Dark)" data-height="900" scrolling="no" allowtransparency="true" loading="lazy" style="width:100%;height:350px;border:none;border-radius:10px;background:transparent;overflow:hidden" ></iframe> </div> <!-- Provider script (only once) --> <script src="https://updates.cyberleadhub.com/js/form_embed.js"></script>