有理数指数的幂法则

How can we find the derivative of $x^{\frac{p}{q}}$, where p, q are positive integers?

We let $p=qr$ to keep things simple.

Intuitively, the answer should be $rx^{r-1}$, because this is the answer when $r$ is an integer.

We can use the binomial theorem here, too!

** Generalized Binomial Theorem ** This is the binomial theorem generalized when the exponent isn't an integer.

Basically we redefine $\binom{n}{k}$ when n is not an integer to mean $\frac{n(n-1)...(n-k+1)}{k!}$.

By redefining binomial coefficients like this, we obtain $(x+y)^r = x^r + r x^{r-1} y + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + ...$.

Notice that this is an **infinite series **. When r is an integer, we will eventually have a $r-r=0$ factor, so the series isn't infinite. However, when r is not an integer, this doesn't happen.

Anyway, applying the generalized binomial theorem, we find

$(x+h)^r = x^r + r x^{r-1} h + \frac{r(r-1)}{2!} x^{r-2}{h^2} + ..$

Then $\frac{(x+h)^r - x^r}{h}$ becomes

$rx^{r-1} + \frac{r(r-1)}{2!} x^{r-2} h + \frac{r(r-1)(r-2)}{3!} x^{r-3} h^2$ + ...

We want the limit of this as h approaches 0, so we are left with $\boxed{rx^{r-1}}$.

So we have shown the derivative of $x^r$ is $rx^{r-1}$, not just for integers but for all rationals.

Now we know that $\frac{d}{dx} x^n = n x^{n-1}$ when $n$ is a positive integer. Is the same thing true for other numbers?

With some work, we can extend this rule to the case when $n$ is a rational number: $n = \frac{p}{q}$ where $p$ and $q$ are integers. Here are three ways to do it.

Using implicit differentiation

If $y = x^{p/q}$, then $y^q = x^p$. We can differentiate both sides to get $$q y^{q-1} \frac{dy}{dx} = p x^{p-1}.$$ Solving for $\frac{dy}{dx}$ and then substituting back $y = x^{p/q}$ yields:

Using the chain rule

Let $f(x) = x^{p/q}$, $g(x) = x^q$, and $h(x) = x^p$. We've set this up so that $g(f(x)) = h(x)$. We also know that $g'(x) = q x^{q-1}$ and $h'(x) = p x^{p-1}$, but we don't (yet) know $f'(x)$.

By using the chain rule, we get:

Note that we're using the chain rule on the composition $g(f(x))$ when we don't yet know that $f$ has a derivative. So technically, all this approach tells us is that if $f(x)$ is differentiable, this must be its derivative.

Using the limit definition

This approach is exceptionally painful, but it can be done. To make things slightly simpler, here is a way to do it for $f(x) = \sqrt{x} = x^{1/2}$. $$\begin{align*} f'(x) &= \lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \\ &= \lim_{h\to 0} \frac{(\sqrt{x+h} - \sqrt{x}) \cdot (\sqrt{x+h} + \sqrt{x})}{h \cdot (\sqrt{x+h} + \sqrt{x})} \\ &= \lim_{h\to 0} \frac{(x+h) - x}{h \cdot (\sqrt{x+h} + \sqrt{x})} \\ &= \lim_{h\to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2 \sqrt{x}}. \end{align*}$$ So $f'(x) = \frac1{2 \sqrt{x}} = \frac{1}{2} \cdot x^{-1/2}.$

This approach used the "difference of squares" factorization $a^2 - b^2 = (a+b)(a-b)$, with $a = \sqrt{x+h}$ and $b = \sqrt{x}$, to convert the numerator into the form $(x+h) - x = h$. To differentiate $x^{1/q}$, we could use the same approach with the factorization $$a^q - b^q = (a - b)(a^{q-1} + a^{q-2} b + \cdots + a b^{q-2} + b^{q-1}).$$ Then the chain rule (applied in the usual direction this time) tells us the derivative of $x^{p/q} = \left(x^{1/q}\right)^p$.

For $f(x)=x^\frac pq$, we can use logarithmic differentiation to find $\frac d{dx}(f(x))$: $$\frac d{dx}(\ln(f(x))) = \frac d{dx}\left(\frac pq\ln(x)\right)$$ Using the chain rule, we get $$\frac1{f(x)} \cdot \frac d{dx}(f(x)) = \frac pq \cdot \frac1x$$

Simplifying, $$\frac d{dx}(f(x)) = \frac pq\cdot\frac1x\cdot f(x)$$ $$\frac d{dx}(f(x)) = \frac pq\cdot\frac{x^\frac pq}x$$ $$\therefore \frac d{dx}(x^\frac pq) = \frac pq x^{\frac pq-1}$$