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So, I'm aware that proving $\displaystyle \frac{d}{dx}x^2$ or higher degrees can be proven by first principles ($\displaystyle{\lim_{dx\to0}}\frac{f(x+dx)-f(x)}{dx}$) but how can $\displaystyle \frac{d}{dx}x^n$ be proven? Where $n$ is obviously any degree of $x$.

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  • $\begingroup$ Can you clarify which first principles you want to avoid? $\endgroup$ – Chris Culter Aug 23 '17 at 8:59
  • $\begingroup$ @ChrisCulter I didn't actually want to avoid them, I wanted to know whether or not there was a way of proving $\displaystyle \frac{d}{dx}x^n$ without substituting values in for $n$ $\endgroup$ – joshuaheckroodt Aug 23 '17 at 9:04
  • $\begingroup$ What are first principles? $\endgroup$ – user297841 Aug 23 '17 at 9:20
  • $\begingroup$ @satoukibi $\displaystyle {\lim_{dx\to0}} \frac{f(x+dx)-f(x)}{dx}$ $\endgroup$ – joshuaheckroodt Aug 23 '17 at 9:22
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    $\begingroup$ For a proof in the case where $n$ is any real number, see Emilio Novati's answer to this question: math.stackexchange.com/questions/1335500/… $\endgroup$ – ekkilop Aug 23 '17 at 9:25
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To be honest, the most explanatory proof comes from "first principals", but includes the use of the binomial theorem, as well. The following is a rigorous proof:

$\displaystyle {\lim_{dx\to0}} \frac{(x+dx)^n-x^n}{dx} $

and from here, the binomial theorem can be used.

$\displaystyle {\lim_{dx\to0}} \frac{\sum_{k=0}^n{n\choose k}x^{n-k}dx^k - x^n}{dx}$

$\displaystyle {\lim_{dx\to0}} \frac{{n\choose0}x^n+{n\choose1}x^{n-1}dx+{n\choose2}x^{n-2}dx^2+\ldots+{n\choose n}dx^n-x^n}{dx}$

$\displaystyle {\lim_{dx\to0}}\frac{x^n+nx^{n-1}dx+{n \choose 2}x^{n-2}dx^2+\ldots+dx^n-x^n}{dx}$

$\displaystyle {\lim_{dx\to0}}\frac{nx^{n-1}dx+{n \choose 2}x^{n-2}dx^2+\ldots+dx^n}{dx}$

$\displaystyle {\lim_{dx\to0}}nx^{n-1}+{n\choose2}x^{n-2}dx+\ldots+dx^{n-1}$

now take the limit:

$\displaystyle {\lim_{dx\to0}}nx^{n-1}+{n\choose2}x^{n-2}dx+\ldots+dx^{n-1}=nx^{n-1}$

and we know that all terms after $xn^{n-1}$ will become $0$ when the limit is taken as their degree of $dx$ always exceeds $1$.

QED

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Proof by induction, $n \in \mathbb{N}$.

$n=1:$ $\dfrac{dx}{dx} = 1$;

Assume $ \dfrac{dx^n}{dx} = nx^{n-1}$.

Step for $n+1$ using the product rule:

$\dfrac{d}{dx} x^{n+1} = \dfrac{d}{dx} ( x x^n) =$

$\dfrac{dx}{dx} (x^n) + x \dfrac{dx^n}{dx} =$

$x^n + x (nx^{n-1}) = (n+1)x^n.$

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It's not altogether clear which "principles" you want to avoid using.

Here are some arguments:

  • $(1)$ \begin{align} f'(x) & = \lim_{t\to x} \frac{t^n - x^n}{t-x} = \lim_{t\to x} \frac{(t-x)(t^{n-1} + t^{n-2} x + t^{n-3}x^2 + t^{n-4} x^3 + \cdots+ x^{n-1})}{t-x} \\[10pt] & = \lim_{t\to x} (t^{n-1} + t^{n-2} x + t^{n-3}x^2 + t^{n-4} x^3 + \cdots+ x^{n-1}) \\[10pt] & = x^{n-1} + x^{n-1} + x^{n-1} + \cdots + x^{n-1} = nx^{n-1}. \end{align}

  • $(2)$ First it is easy to prove $\dfrac d {dx} x^1 = 1x^0.$ Then use the product rule and mathematical induction: \begin{align} \frac d {dx} x^n = \frac d{dx} (x \cdot x^{n-1}) & = x \frac d {dx} x^{n-1} + x^{n-1} \frac d {dx} x \\[10pt] & = x \cdot (n-1)x^{n-2} + x^{n-1} \cdot 1 \\[10pt] & = (n-1)x^{n-1} + x^{n-1} = nx^{n-1}. \end{align}

  • $(3)$ Consider the cube $[0,x]^n \subseteq [0,\infty)^n.$

    As $x$ changes, $n$ of the $2n$ faces of this cube move.

    The amount by which each face moves is the amount by which $x$ changes; therefore the rate at which each face moves is the rate at which $x$ changes.

    Each face has size $x^{n-1}.$

    Then use this: \begin{align} & \Big( \text{size of the boundary} \Big) \times \Big( \text{rate of motion of the boundary} \Big) \\[10pt] = {} & \Big( \text{rate of change of size of the bounded region} \Big). \end{align} The size of the bounded region is $x^n$. Hence $d(x^n) = nx^{n-1} \, dx.$

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