Positive rational powers

What about functions like $f:[0,\infty)\rightarrow[0,\infty)$, $f(x) = \sqrt{x}$ or $f:[0,\infty)\rightarrow[0,\infty)$, f(x) = x^5/3? Here we have to be a lot cleverer with our algebra, but remarkably, the results are identical to the case of integer powers, with the exception of the slope at (0,0) in some cases. The reason things work out is the following clever observation. Suppose that p is a positive integer and x and y are non-negative numbers. Then

x - y = (x^1/p)^p - (y^1/p)^p

so (x-y) is a difference of $p^{\rm th}$ powers of the $p^{\rm th}$ roots of x and y. So, for example, if $f(x) = \sqrt{x}$ then (remembering $a \geq 0$ and $a+h\geq 0$)

$$\begin{array} {rcl} {\displaystyle \frac{f(a+h)-f(a)}{a+h-a}... ... & = & {\displaystyle \frac{1}{\sqrt{a+h}+\sqrt{a}}}\end{array}$$

so if a > 0 then it appears that as h approaches the slopes of chords approach

$$\frac{1}{2\sqrt{a}} = a^{-1/2}/2 = (1/2)a^{(1/2)-1}$$

which fits the pattern of f(x) = x^p giving f'(a) = px^p-1 which we observed for p an integer.

If a = 0 then

$$\begin{array} {rcl} {\displaystyle \frac{f(0+h)-f(0)}{a+h-a}... ...t{h}}{h}}\\ & = & {\displaystyle \frac{1}{\sqrt{h}}}\end{array}$$

The result of letting h approach is to get numbers which are larger than any fixed real number, and our method fails. We shall see later that there is still a line deserving of the name tangent line, but it is vertical and has no slope.

Here is the same idea applied to $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$, $f(x) = \sqrt[3]{x}$. This time we will get a difference of cubes! For the moment, let x= a+h so things don't get even messier than they will be.

$$\begin{array} {rcl} {\displaystyle \frac{f(x) - f(a)}{x-a} }... ...]{x})^2 + \sqrt[3]{x}\sqrt[3]{a} + (\sqrt[3]{a})^2}}\end{array}$$

As h approaches we see that x approaches a and, so the slopes of the chords appear to approach

$$\frac{1}{3(\sqrt[3]{a})^2} = (1/3)a^{-2/3} = (1/3)a^{(1/3)-1}$$

(provided $a\neq 0$). If a = 0, the analysis fails (for even more complicated reasons then with square root), but otherwise we see the same pattern as before: to compute f', reduce the exponent by 1 and multiply by the old exponent. In fact, this holds for any positive root at all: if f(x) = x^1/p and $a\neq 0$ then f'(a) = (1/p)a^(1/p)-1.