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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 $x$.

joshuaheckroodt
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  • Can you clarify which first principles you want to avoid? – Chris Culter Aug 23 '17 at 08:59
  • @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 $n$ – joshuaheckroodt Aug 23 '17 at 09:04
  • What are first principles? –  Aug 23 '17 at 09:20
  • @satoukibi $\displaystyle {\lim_{dx\to0}} \frac{f(x+dx)-f(x)}{dx}$ – joshuaheckroodt Aug 23 '17 at 09:22
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    For a proof in the case where $n$ is any real number, see Emilio Novati's answer to this question: https://math.stackexchange.com/questions/1335500/proof-of-the-derivative-of-xn – ekkilop Aug 23 '17 at 09:25
  • You can't avoid the first principles (ie definition of derivative). The rules of differentiation are consequences of the definition. So even if you are using product/chain rule, you are in effect using using the first principles although in an implicit manner. Normally some textbooks ask you to evaluate the derivative without the use of those rules and just via definitions. The other way round is not possible. In simpler terms, you can't add two numbers without using the definition of "addition of two nunbers". – Paramanand Singh Aug 24 '17 at 04:50

3 Answers3

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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

joshuaheckroodt
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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:

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

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

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

Peter Szilas
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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 $d(x^n) = nx^{n-1} \, dx.$

Michael Hardy
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