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So I was thinking about what I have learnt and I realised that I kind of took the derivative of a function for granted. So I did some research as I wanted to find out how this was discovered and I stumbled upon this.

More specially, here is a passage from it:

Without going into too much complicated detail, Newton (and his contemporary Gottfried Leibniz independently) calculated a derivative function $f'(x)$ which gives the slope at any point of a function $f(x)$. This process of calculating the slope or derivative of a curve or function is called differential calculus or differentiation (or, in Newton’s terminology, the “method of fluxions” - he called the instantaneous rate of change at a particular point on a curve the "fluxion", and the changing values of x and y the "fluents"). For instance, the derivative of a straight line of the type $f(x) = 4x$ is just $4$; the derivative of a squared function $f(x) = x^2$ is $2x$; the derivative of cubic function $f(x) = x^3$ is $3x^2$.

I was wondering if anyone could explain (or point me to a resource) the "complicated details" (or hopefully, a rigorous proof) of how the derivative function $f'(x)$ was discovered?

Michael Hardy
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YellowPillow
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1 Answers1

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When I introduce the derivative in my first semester calculus, I show my students the following relatively straightforward proof of the so-called "power rule".

Notice primarily that for all positive integers $n$ and all real numbers $a$ we have:

\begin{align} (x-a)\sum_{i=0}^{n-1}x^ia^{n-1-i} &=x\sum_{i=0}^{n-1}x^ia^{n-1-i}-a\sum_{i=0}^{n-1}x^ia^{n-1-i} \\ &=\sum_{i=0}^{n-1}x^{i+1}a^{n-1-i}-\sum_{i=0}^{n-1}x^ia^{n-1-i+1} \\ &=x^{(n-1)+1}a^{n-1-(n-1)}+\sum_{i=0}^{n-2}x^{i+1}a^{n-1-i}-\left(\sum_{i=1}^{n-1}x^ia^{n-i}+x^0a^{n-0}\right) \\ &=x^na^0+\sum_{i=1}^{n-1}x^ia^{n-i}-\sum_{i=1}^{n-1}x^ia^{n-i}-x^0a^n \\ &=x^n-a^n \end{align}

Now remember that one way of defining the derivative of function $f(x)$ at $x=a$ is given by the following limit provided it exists:

$$f'(a):=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$

Now we'll use the above definition of the derivative together with our work at the beginning to derive the function $f(x)=x^n$ at $x=a$:

\begin{align*} f'(a)&=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} =\lim_{x\to a}\frac{x^n-a^n}{x-a} \\ &=\lim_{x\to a}\frac{(x-a)\sum_{i=0}^{n-1}x^ia^{n-1-i}}{x-a} =\lim_{x\to a}\sum_{i=0}^{n-1}x^ia^{n-1-i} \\ &=\sum_{i=0}^{n-1}a^ia^{n-1-i} =\sum_{i=0}^{n-1}a^{i+n-1-i} =\sum_{i=0}^{n-1}a^{n-1} \\ &=na^{n-1} \end{align*}

In particular for every real number $a$ we have $f'(a)=na^{n-1}$. Hence as a function $\frac{d(x^n)}{dx}=f'(x)=nx^{n-1}$ as we wanted.

Note: Historically a similar proof was used, but that proof relies on "infinitely small numbers," a concept which most mathematicians agree was not well-founded at the time. This proof is considered rigorous, although I have left some words out (you should be able to fill them in as long as you know your limit laws!).

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