> [!NOTE] Lemma (The derivatives of monomials with integer exponents) > If $n\in \mathbb{N}$ then the [[Fréchet Differentiation|derivative]] of $x\mapsto x^{n}$ is $x\mapsto nx^{n-1}$. *Proof*. If $f(x)=x$ then for every $c\in\mathbb{R}$ $f'(c) = \lim_{ h \to 0 } \frac{f(c+h)-f(c)}{h} = \lim_{ h \to 0 } \frac{c+h-c}{h} = 1. $Assume inductively that we have the result for $f(x)=x^{n}$. Then $x^{n+1}=xf(x)$. Differentiating both sides w.r.t. $x$ using [[Derivative of Sum of Differentiable Real Functions|product rule]] gives: $f(x)+xf'(x)=x^{n}+xnx^{n-1} = (n+1)x^{n}.$ *Proof*. Note that $\frac{y^{n} - x^{n}}{y-x} = y^{n-1} + y^{n-2}x +\dots +x^{n-1}$which is polynomial in $y$ hence it must be [[Algebra of Continuous Real Functions|continuous]] at $y\in\mathbb{R}$. Also its value at $x$ is $nx^{n-1}$. So $\frac{d}{dx} (x^{n}) = \lim_{ y \to x } \frac{y^{n}-x^{n}}{y-x} = nx^{n-1} $ > [!NOTE] Lemma (The derivative of monomials) > If $n\in\mathbb{R}$, then the derivative of $x \mapsto x^{n}$ is $x \mapsto nx^{n-1}$. *Proof*. Note that $x^{n}$ is [[Real Power of Real Number|defined]] as $\exp(p\log x)$. Using [[Derivative of Real Exponential Function|the derivative of the exponential]] $\frac{d}{dx} (x^{n}) = \frac{p}{x} \cdot \exp(p\log x) = \frac{p}{x} x^{p} = p x^{-1} $as required.