整数指数的幂法则

Power rule for integer exponents proved in 2 methods

One of the most basic differentiation formulas is $$\frac{d}{dx} x^n = n x^{n-1}.$$ (For example, $\frac{d}{dx} x^1 = 1$.) Since derivatives distribute across sums (from linearity of derivatives), and the derivative of any constant is zero, this lets us differentiate any polynomial. For example, $$\begin{align*} \frac{d}{dx}(10x^4 + 2x + 1) &= \frac{d}{dx}(10x^4) + \frac{d}{dx}(2x) + \frac{d}{dx}(1) \\ &= 10(4x^3) + 2(1) + 0 \\ &= 40x^3 + 2. \end{align*}$$ The formula works for any value of $n$ for which we can even say what $x^n$ means. We'll begin by proving the formula when $n$ is a positive integer, in two ways.

Method 1: The product rule

We already know that $\frac{d}{dx} x^1 = 1$ and $\frac{d}{dx} x^0 = 0$. So the first place to start is $x^2$, which we can conveniently write as $x \cdot x$.

By the product rule, $$\frac{d}{dx}(x \cdot x) = \frac{d}{dx}(x) \cdot x + x \cdot \frac{d}{dx}(x) = 1 \cdot x + x \cdot 1 = 2x.$$ So the formula $\frac{d}{dx} x^n = n x^{n-1}$ checks out when $n=2$.

What about $n=3$, the next case? We can write $x^3$ as $x \cdot x^2$, and we know the derivative of both $x$ and $x^2$. By using the product rule again, we get $$\frac{d}{dx}(x \cdot x^2) = \frac{d}{dx}(x) \cdot x^2 + x \cdot \frac{d}{dx}(x^2) = 1 \cdot x^2 + x \cdot 2x = 3x^2.$$ So the formula $\frac{d}{dx} x^n = n x^{n-1}$ checks out when $n=3$.

We could also find the derivative of $x^4$ in almost exactly the same way... but if we have to write a separate proof for each value of $n$, we'll be at it for a while. So let's just write $x^n$ as $x \cdot x^{n-1}$ and do the general case: $$\begin{align*} \frac{d}{dx}(x^n) &= \frac{d}{dx}(x \cdot x^{n-1}) \\ &= \frac{d}{dx}(x) \cdot x^{n-1} + x \cdot \frac{d}{dx}(x^{n-1}) \\ &= 1 \cdot x^{n-1} + x \cdot (n-1)x^{n-1} \\ &= nx^{n-1}. \end{align*}$$ Is it okay that we used the formula to find $\frac{d}{dx} x^{n-1}$ when we were deriving $\frac{d}{dx} x^n$? Normally, that would be a little sketchy. In this case, it's okay, because to prove that the formula works for $n$, we use the fact that it works for $n-1$. For example, to prove that $\frac{d}{dx}(x^{100}) = 100 x^{99}$, we need to know that the formula works for $n=99$... which requires the formula for $n=98$... which uses the formula for $n=97$... and eventually we get down to $n=1$, which we know.

This "bootstrapping" technique is known as a proof by induction.

Method 2: The Binomial Theorem

We can also try applying the limit definition directly. For example, to find the derivative of $x^2$, we find $$\begin{align*} \frac{d}{dx}(x^2) &= \lim_{h\to 0} \frac{(x+h)^2 - x^2}{h} \\ &= \lim_{h\to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\ &= \lim_{h\to 0} 2x+h = 2x. \end{align*}$$ All we needed to know was that $(x+h)^2$ expands as $x^2 + 2xh + h^2$.

To differentiate $x^n$, we will need to expand $(x + h)^n$, instead. The Binomial Theorem tells us that if we do, we'll get $$(x+h)^n = {n \choose 0}x^n + {n\choose 1}x^{n-1}h + {n \choose 2}x^{n-2} h^2 + \cdots + {n \choose n}h^n,$$ where ${n \choose k}$ is the binomial coefficient, equal to $\frac{n!}{k!(n-k)!}$.

We don't really need to know that, though. All we need to know is that $$(x+h)^n = x^n + n x^{n-1} h + P(x, h)\cdot h^2$$ where $P(x, h)$ is some polynomial in $x$ and $h$. The coefficient of $x^n$ is $1$ (we only get $x^n$ by choosing $x$ from every $(x+h)$ factor) and the coefficient of $x^{n-1} h$ is $n$ (there are $n$ ways to get $x^{n-1}h$: choose $h$ from one of the $(x+h)$ factors, and $x$ from the rest). Since ${n \choose 0} = 1$ and ${n\choose 1} = n$, the Binomial Theorem supports us.

Therefore $$\begin{align*} \frac{d}{dx}(x^n) &= \lim_{h\to 0} \frac{(x+h)^n - x^n}{h} \\ &= \lim_{h\to 0} \frac{x^n + n x^{n-1} h + P(x,h)\cdot h^2 - x^n}{h} \\ &= \lim_{h\to 0} n x^{n-1} + P(x,h) \cdot h \\ &= n x^{n-1}. \end{align*}$$

Negative exponents

Everything we've done so far only works when $n \ge 0$. What about $x^{-1} = \frac{1}{x}$ or $x^{-2} = \frac{1}{x^2}$?

We can show that the formula works here as well by writing $x^{-n}$ as $\frac{1}{x^n}$ and applying the quotient rule: $$\begin{align*} \frac{d}{dx} x^{-n} &= \frac{d}{dx}\left(\frac{1}{x^n}\right) \\ &= \frac{\frac{d}{dx}(1) \cdot x^n - 1 \cdot \frac{d}{dx}(x^n)}{(x^n)^2} \\ &= \frac{0 - n x^{n-1}}{x^{2n}} \\ &= - n x^{-n-1}. \end{align*}$$

How can we find the derivative of $f(x)=x^n$ , where n is an integer?

We'll need to take a look at the definition of a derivative- It is the limit as h approaches 0 of $\frac{f(x+h)-f(x)}{h}$.

We plug in $f(x)=x^n$ to get $\frac{(x+h)^n-x^n}{h}$. We can't just plug in $h=0$ because the fraction would become $\frac{0}{0}$.

By expanding the numerator, we obtain that the numerator equals $x^n + \binom{n}{1} x^{n-1} h + ... + h^n - x^n$.

So the numerator equals $\binom{n}{1} x^{n-1} h + ... + h^{n-1} x \binom{n}{n-1} + h^n$.

The denominator of the fraction is h, so we divide by h.

The derivative then becomes the limit, as h approaches 0, of $\binom{n}{1} x^{n-1} + \binom{n}{2} x^{n-2} h + ... + h^{n-1}$.

We plug in $h=0$, so the expression becomes $nx^{n-1}$.

Therefore, $(x^n)' = nx^{n-1}$.

This makes lots of sense too- The derivative of a constant becomes 0, and the derivative of $f(x)=ax+b$ is $a$, the slope of the line.

How can we find the derivative of $a^x$ (with respect to a)?

First we take a look at the definition of a derivative: It is the limit as $h$ approaches $0$ of $\frac{f(a+h)-f(a)}{h}$. We can't just plug in $h=0$ right now, because then the expression becomes $\frac{0}{0}$.

Now we let $f(a)=a^x$.

Plugging it into the formula, we find the derivative of $a^x$ is $\frac{(a+h)^x - a^x}{h}$.

We use binomial expansion on the numerator- it becomes $a^x + a^{x-1} h \binom{x}{1} + ... + ah^{x-1} \binom{x}{x-1} + h^x - a^x$.

Therefore, $\frac{f(a+h)-f(a)}{h} = a^{x-1} \binom{x}{1} + a^{x-2} h \binom{x}{2} + ... + h^x$.

But remember, we want to find the value of this as h approaches 0! Then we can plug in $h=0$ so that everything reduces to $xa^{x-1}$.

We have shown that the derivative of $a^x$ is $xa^{x-1}$.

This makes lots of sense: For example, the derivative of any constant like 5 becomes $0.$ Meanwhile, the derivative of $f(x)=ax+b$ is $a$, which is the slope of the line.