> [!summary] This proof uses definitions of a limit and derivative to prove the chain rule. > Key Result: $\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g(x)$ >[!info]+ Read Time **⏱ 2 mins** # Mathematical Proof > [!warning] Assumptions To prove the derivative of a chain of functions 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|limit product rule]] is $\displaystyle \lim_{ x \to a } [f(x)\cdot g(x)]=\displaystyle \lim_{ x \to a } f(x )\cdot \displaystyle \lim_{ x \to a }g(x)$ Prove that $\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g(x)$ $ \begin{array}{c} \frac{d}{dx}f(g(x)) = \displaystyle \lim_{ h \to 0 } \frac{f(g(x+h))-f(g(x))}{h} \\ \\ \text{Multiply top and bottom by $g(x+h)-g(x)$}\\ =\displaystyle \lim_{ h \to 0 } \frac{f(g(x+h))-f(g(x))}{h}\cdot \frac{g(x+h)-g(x)}{g(x+h)-g(x)} \\ \\ \text{Using the limit product rule we can configure to:}\\ \displaystyle \lim_{ h \to 0 } \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\displaystyle \lim_{ h \to 0 } \frac{g(x+h)-g(x)}{h} \\ \\ \displaystyle \lim_{ h \to 0 } \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot g'(x) \\ \\ \text{Let $k=g(x+h)-g(x)$} \\ \text{By rearanged the let statment}\\ g(x)+k = g(x+h) \\ \\ \\ \text{As $h\to 0, k \to 0$, so now:} \\ \displaystyle \lim_{ k \to 0 } \frac{f(g(x)+k)-f(g(x))}{k}\cdot g'(x) \\ \\ \text{By definition of a limit this is }\\ f'(g(x))\cdot g'(x) \end{array} $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/m0LZX19DyyI?si=pnVk0nhIprGkVrg4" 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). ---