Applying the Power Rule · 应用幂法则
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| power/ˈpaʊə/ | 幂 | mì |
| Power Rule/ˈpaʊə ruːl/ | 幂法则 | mì fǎ zé |
| integer/ˈɪntɪdʒə/ | 整数 | zhěng shù |
A shortcut that replaces the limit
- Computing every derivative from the limit definition is slow. There's a pattern.
- For powers of $x$, one rule does it instantly: the Power Rule 幂法则.
-
$$\frac{d}{dx}\big[x^n\big]=n\,x^{\,n-1}$$
- "Bring the exponent down as a coefficient, then subtract one from the exponent."
一个替代极限的捷径
- 每个导数都用极限定义来算太慢了。其实有规律。
- 对 $x$ 的幂,一条规则瞬间搞定:幂法则。
-
$$\frac{d}{dx}\big[x^n\big]=n\,x^{\,n-1}$$
- "把指数当系数拿下来,再把指数减一。"
A power and its tangent slope · 一个幂及其切线斜率
y = x²
For · 支持 $y=x^2$ the power rule gives slope $2x$ — move the point and check the tangent slope matches $2x$. · 对于 $y=x^2$，幂法则给出斜率 $2x$——移动该点并检查切线斜率是否匹配 $2x$。
Watch it work
- $\dfrac{d}{dx}[x^3]=3x^2$ — the $3$ drops in front, the exponent goes $3\to2$.
- $\dfrac{d}{dx}[x]=1\cdot x^0=1$ — the derivative of $x$ is $1$.
- $\dfrac{d}{dx}[x^{10}]=10x^9$.
- It matches the limit definition (we found $\frac{d}{dx}[x^2]=2x$) but takes one line.
看它运作
- $\dfrac{d}{dx}[x^3]=3x^2$——$3$ 落到前面,指数 $3\to2$。
- $\dfrac{d}{dx}[x]=1\cdot x^0=1$——$x$ 的导数是 $1$。
- $\dfrac{d}{dx}[x^{10}]=10x^9$。
- 它与极限定义一致(我们求过 $\frac{d}{dx}[x^2]=2x$),但只需一行。
What is $\dfrac{d}{dx}[x^5]$? · $\dfrac{d}{dx}[x^5]$是什么？
Bring down $5$, reduce exponent: $5x^{5-1}=5x^4$. · 将 $5$ 提下来，指数减一：$5x^{5-1}=5x^4$。
If $f(x)=x^4$, find $f'(2)$. · 若 $f(x)=x^4$，求 $f'(2)$。
$f'(x)=4x^3$, so $f'(2)=4\cdot8=32$. · $f'(x)=4x^3$，所以 $f'(2)=4\cdot8=32$。
It works for any exponent
- $n$ can be a negative or a rational (fractional) exponent, not just a positive integer 整数.
- Negative: $\dfrac{d}{dx}[x^{-2}]=-2x^{-3}$.
- Rational: $\dfrac{d}{dx}\big[x^{1/2}\big]=\tfrac12 x^{-1/2}$.
- Same rule every time — bring down $n$, reduce the exponent by one.
它对任何指数都成立
- $n$ 可以是负指数或有理(分数)指数,不只是正整数。
- 负指数:$\dfrac{d}{dx}[x^{-2}]=-2x^{-3}$。
- 有理指数:$\dfrac{d}{dx}\big[x^{1/2}\big]=\tfrac12 x^{-1/2}$。
- 每次都是同一条规则——把 $n$ 拿下来,指数减一。
$\dfrac{d}{dx}[x^{-2}]=-2x^{\square}$. The exponent $\square$ is . · $\dfrac{d}{dx}[x^{-2}]=-2x^{\square}$。指数 $\square$ 是。
Reduce $-2$ by one: $-2-1=-3$. · 将 $-2$ 的指数减一：$-2-1=-3$。
The Power Rule differentiates which of these directly (after rewriting)? · 重写后，幂法则可以直接对以下哪一项求导？
All are powers of $x$ after rewriting; $5^x$ is an exponential, not a power of $x$. · 重写后它们都是 $x$ 的幂；$5^x$ 是指数函数，不是 $x$ 的幂。
Rewrite first, then differentiate
- Radicals and reciprocals hide powers — rewrite them as $x^n$ before applying the rule.
- $\sqrt{x}=x^{1/2}$, so $\dfrac{d}{dx}[\sqrt x]=\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt x}$.
- $\dfrac{1}{x^3}=x^{-3}$, so $\dfrac{d}{dx}\!\left[\dfrac1{x^3}\right]=-3x^{-4}=-\dfrac{3}{x^4}$.
- Turning every term into a power is the key setup step.
先改写,再求导
- 根号与倒数藏着幂——先把它们写成 $x^n$,再套用规则。
- $\sqrt{x}=x^{1/2}$,所以 $\dfrac{d}{dx}[\sqrt x]=\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt x}$。
- $\dfrac{1}{x^3}=x^{-3}$,所以 $\dfrac{d}{dx}\!\left[\dfrac1{x^3}\right]=-3x^{-4}=-\dfrac{3}{x^4}$。
- 把每一项都变成幂是关键的准备步骤。
Rewrite and differentiate: $\dfrac{d}{dx}[\sqrt x]$ equals... · 重写并求导：$\dfrac{d}{dx}[\sqrt x]$ 等于……
$\sqrt x=x^{1/2}\Rightarrow \tfrac12 x^{-1/2}=\tfrac1{2\sqrt x}$.
The Power Rule applies to $2^x$, giving $x\cdot 2^{x-1}$. · 幂法则适用于 $2^x$，得出 $x\cdot 2^{x-1}$。
The Power Rule needs a variable base with constant exponent; $2^x$ is an exponential (next lesson). · 幂法则需要一个变量底数和常数指数；$2^x$ 是一个指数函数（下节课内容）。
The Power Rule is for a variable base with a constant exponent ($x^n$). It does not apply to a constant base with a variable exponent — $\frac{d}{dx}[2^x]\neq x\,2^{x-1}$. (That is an exponential, handled in the next lesson.) And don't forget to reduce the exponent: $\frac{d}{dx}[x^{-2}]=-2x^{-3}$, not $-2x^{-2}$.
幂法则适用于变量底、常数指数($x^n$)。它不适用于常数底、变量指数——$\frac{d}{dx}[2^x]\neq x\,2^{x-1}$。(那是指数函数,下一课处理。)而且别忘了指数要减一:$\frac{d}{dx}[x^{-2}]=-2x^{-3}$,不是 $-2x^{-2}$。
Differentiate $g(x)=\sqrt[3]{x^2}+\dfrac{1}{x}$.
- Rewrite as powers: $g(x)=x^{2/3}+x^{-1}$.
- Power rule term by term: $\tfrac23 x^{-1/3}$ and $-1\cdot x^{-2}$.
- $g'(x)=\tfrac23 x^{-1/3}-x^{-2}=\dfrac{2}{3\sqrt[3]{x}}-\dfrac1{x^2}$.
对 $g(x)=\sqrt[3]{x^2}+\dfrac{1}{x}$ 求导。
- 改写成幂:$g(x)=x^{2/3}+x^{-1}$。
- 逐项用幂法则:$\tfrac23 x^{-1/3}$ 与 $-1\cdot x^{-2}$。
- $g'(x)=\tfrac23 x^{-1/3}-x^{-2}=\dfrac{2}{3\sqrt[3]{x}}-\dfrac1{x^2}$。
The Power Rule $\frac{d}{dx}[x^n]=n\,x^{n-1}$ works for any constant exponent — integer, negative, or rational. Rewrite radicals and reciprocals as powers first, then bring the exponent down and reduce it by one. It replaces the limit definition for power functions.
幂法则 $\frac{d}{dx}[x^n]=n\,x^{n-1}$ 对任何常数指数都成立——整数、负数或有理数。先把根号与倒数改写成幂,再把指数拿下来并减一。对幂函数,它替代了极限定义。