$ \huge \begin{align*} y &= x^n \\ y' &= nx^{n-1} \end{align*}$ ## Proof / Reasoning ### Difference between 2 $n$th powers rule If $a$ and $b$ are real numbers, and $n$ is a positive integer, then $ a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2} b+a^{n-3} b^2+\ldots+a b^{n-2}+b^{n-1}\right) $ ### Proof $\begin{align*} f(x) &= x^n \\ \frac{df}{dx}(x) &= \lim_{h\to 0}\frac{f(x+h) - f(x)}{h} \\ &= \lim_{h\to 0}\frac{(x+h)^n - x^n}{h} \\ \frac{df}{dx}(x) &= \lim _{h \rightarrow 0} \frac{((x+h)-x)\left((x+h)^{n-1}+(x+h)^{n-2} x+\ldots+x^{n-1})\right)}{h}\\ & =\lim _{h \rightarrow 0} \frac{h\left[(x+h)^{n-1}+(x+h)^{n-2} x+\ldots+x^{n-1}\right]}{h} \\ & =\lim _{h \rightarrow 0}\left[(x+h)^{n-1}+(x+h)^{n-2} x+\ldots+x^{n-1}\right] \\ &= x^{n-1}+x^{n-1}+\ldots+x^{n-1}\\ \frac{df}{dx}(x) &= nx^{n-1} \end{align*}$