Using the binomial theorem to generate a geometric proof of the derivative.

Question

According to wikipedia, if we wanted to prove $$(x^n)'=nx^{n-1}$$ geometrically by creating an $n$-dimensional hypercube $$(x+\Delta x)^n$$ and setting $a=x$ and $b=\Delta x$, we could expand using the binomial theorem to $$x^n + nx^{n-1}\Delta x + \binom{n}{2}x^{n-2}(\Delta x)^2 + ...$$ (and this is where I lose it) the terms $(\Delta x)^2$ and higher become negligible as $\Delta x \to 0$, which yields the formula $$(x^n)'=nx^{n-1}$$ However, if $\Delta x \to 0$, wouldn't that yield the formula $$(x^n)'=x^n$$ as every term but the first in the expanded version is reduced to $0$ by being multiplied by $(\Delta x)^k$, where $1\le k \le n$?

I don't understand where the leading $x^n$ goes, or why $nx^{n-1}\Delta x$ does not equal 0 as $\Delta x \to 0$

Answer 1

To take the derivative, you need to look at the difference quotient for the power function, not the power function itself: $$ \frac{(x + \Delta x)^{n} - x^{n}}{\Delta x} = nx^{n-1} + \binom{n}{2} x^{n-2}\, \Delta x + \dots. $$ Now letting $\Delta x \to 0$ has the advertised effect.

Answer 2

if you want a geometrical argument, a simple diagram shows that: $$ D(xf) = f + xDf $$ so $$ D(x^n)=D(x.x^{n-1})= x^{n-1}+xD(x.x^{n-2}) \\ =2x^{n-1}+x^2D(x.x^{n-2})= \cdots =nx^{n-1} $$