> [!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 <iframe width="560" height="315" src="https://www.youtube.com/embed/kNhYfmC_UPU?si=R71WzhalQsOMRbOG" 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> --- > 📂 Want to see more structured notes like these? > Help grow the project by [starring Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidian-MathMatter). ---