> [!summary] This proof uses definitions of derivatives and the binomial theorem to prove the power rule. > Key result: $\frac{d}{dx}x^n = nx^{n-1}$ >[!info]+ Read Time **⏱ 1 min** # Mathematical Proof > [!warning] Assumptions To prove the power rule of [[Derivative|derivatives]] through [[Direct Proof|direct proof]], assume the following: > - The the expansion of $(x+y)^n$ using the [[Binomial Theorem|binomial theorem]] is $(x+y)^n = x^n + \binom{n}{1}x^{n-1}y^1+ \binom{n}{2}x^{n-2}y^2+\dots +y^n$ > - The definition of a derivative is $\frac{dy}{dx} = f'(x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h}$ $ \begin{array}{c} \text{Prove that $\frac{d}{dx}x^n = nx^{n-1}$ }\\ \\ \text{From the definition of a derivative}\\ \frac{d}{dx}x^n = \displaystyle \lim_{ h \to 0 } \frac{(x+h)^n -x^n}{h} \\ \\ (x+h)^n = x^n+ nx^{n-1}h+\binom{n}{2}x^{n-2}h^2 + \dots +h^n \\ \\ \begin{align*} \displaystyle \lim_{ h \to 0 } \frac{(x+h)^n -x^n}{h} &= \displaystyle \lim_{ h \to 0 } \frac{ x^n+ nx^{n-1}h+\binom{n}{2}x^{n-2}h^2 + \dots +h^n+x^n}{h} \\ &= \displaystyle \lim_{ h \to 0 } \frac{ \cancel{x^n}+ nx^{n-1}h+\binom{n}{2}x^{n-2}h^2 + \dots +h^n+\cancel{x^n}}{h} \\ &= \displaystyle \lim_{ h \to 0 } \frac{\cancel{h}(nx^{n-1} + \binom{n}{2}x^{n-2}h + \dots +h^{n-1} }{\cancel{h}}\\ &= \displaystyle \lim_{ h \to 0 } (nx^{n-1} + \binom{n}{2}x^{n-2}h + \dots +h^{n-1}) \\ &= nx^{n-1} \end{align*} \\ \\ \\ \text{Therefore from our definition of a derivative} \\ \frac{d}{dx}x^n = nx^{n-1} \end{array} $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/JjOndio6-g4?si=G5rGkkA7aJy6ZsQt" 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> --- > 🧪 Think this could help someone else? [Star Math & Matter on Github](https://github.com/rajeevphysics/Obsidian-MathMatter) to help more learners discover it. ---