The Product Rule · 乘积法则
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Product Rule/ˈprɒdʌkt ruːl/ | 乘积法则 | chéng jī fǎ zé |
Why products need their own rule
- Tempting but wrong: the derivative of a product is not the product of the derivatives.
- $\dfrac{d}{dx}[x\cdot x]=\dfrac{d}{dx}[x^2]=2x$, but $f'g'=1\cdot1=1$. They disagree.
- Products get their own tool: the Product Rule 乘积法则.
- It accounts for both factors changing at once.
为何乘积需要自己的规则
- 诱人却错误:乘积的导数不是导数的乘积。
- $\dfrac{d}{dx}[x\cdot x]=\dfrac{d}{dx}[x^2]=2x$,但 $f'g'=1\cdot1=1$。两者不符。
- 乘积有自己的工具:乘积法则。
- 它考虑到两个因子同时变化。
The derivative of $f\cdot g$ equals $f'\cdot g'$. · $f\cdot g$ 的导数等于 $f'\cdot g'$。
It is · 它是 $f'g+fg'$, not $f'g'$. · 它是 $f'g+fg'$，而不是 $f'g'$。
The rule
- For a product of two functions:
-
$$\frac{d}{dx}\big[f\,g\big]=f'\,g+f\,g'$$
- In words: "derivative of the first times the second, plus the first times derivative of the second."
- Each term differentiates one factor and leaves the other alone; then you add.
规则
- 对两个函数的乘积:
-
$$\frac{d}{dx}\big[f\,g\big]=f'\,g+f\,g'$$
- 用文字说:"第一个的导数乘第二个,加上第一个乘第二个的导数。"
- 每一项只对一个因子求导,另一个保持不动;然后相加。
The Product Rule states $\dfrac{d}{dx}[fg]=$ · 乘积法则表明 $\dfrac{d}{dx}[fg]=$
Derivative of first times second, plus first times derivative of second. · 第一个函数的导数乘以第二个函数，加上第一个函数乘以第二个函数的导数。
A clean worked pattern
- Differentiate $y=x^2\,e^x$. Let $f=x^2$ and $g=e^x$.
- $f'=2x$, $g'=e^x$.
- $y'=f'g+fg'=2x\,e^x+x^2\,e^x$.
- Factor if asked: $y'=x e^x(2+x)$.
一个干净的范例
- 对 $y=x^2\,e^x$ 求导。设 $f=x^2$,$g=e^x$。
- $f'=2x$,$g'=e^x$。
- $y'=f'g+fg'=2x\,e^x+x^2\,e^x$。
- 若题目要求就提取因式:$y'=x e^x(2+x)$。
A product's changing slope · 一个乘积的变化斜率
y = ax² + bx
A product like $x^2 e^x$ has a slope that mixes both factors — the Product Rule captures both changing at once. · 像 $x^2 e^x$ 这样的乘积具有混合两个因子的斜率——乘积法则同时捕捉两者的变化。
Differentiate $y=x^2 e^x$. · 对 $y=x^2 e^x$ 求导。
$f'g+fg'=2x e^x+x^2 e^x$.
For · 支持 $y=x\sin x$, use the product rule to find $y'$ at $x=0$. (Recall $\sin 0=0$, $\cos 0=1$.) · 对于 $y=x\sin x$，使用乘积法则求 $y'$ 在 $x=0$ 处的值。（记得 $\sin 0=0$，$\cos 0=1$。）
$y'=\sin x+x\cos x$; at $0$: $0+0\cdot1=0$. · $y'=\sin x+x\cos x$；在 $0$ 处：$0+0\cdot1=0$。
When you actually need it
- If a product can be multiplied out first, do that — it may be easier ($x^2\cdot x^3=x^5\Rightarrow5x^4$).
- But when the factors won't simplify (like $x^2\sin x$ or $e^x\ln x$), the rule is required.
- Combine it freely with the power and elementary derivatives for each factor.
- Label $f,f',g,g'$ first — organized bookkeeping prevents mistakes.
何时真正需要它
- 若乘积可以先乘开,就乘开——那可能更简单($x^2\cdot x^3=x^5\Rightarrow5x^4$)。
- 但当因子无法化简(如 $x^2\sin x$ 或 $e^x\ln x$)时,就必须用这条规则。
- 对每个因子自由地配合幂法则与基本导数。
- 先标出 $f,f',g,g'$——有条理的记法能防止出错。
For · 支持 $y=x^2\cdot x^3$, the smartest first move is to... · 对于 $y=x^2\cdot x^3$，最明智的第一步是……
Simplifying to $x^5$ first is easier: $5x^4$. Use the Product Rule only when factors won't combine. · 先简化到 $x^5$ 更容易：$5x^4$。仅在因子无法合并时才使用乘积法则。
In each term of the Product Rule, which is true? · 在乘积法则的每一项中，哪一项是正确的？
One factor per term is differentiated; the two terms are summed. Differentiating both in one term is the classic error. · 每项对一个因子求导；两项相加。在一项中对两个因子都求导是经典错误。
The Product Rule is $f'g+fg'$, not $f'g'$. Both terms are needed, and each keeps the other factor undifferentiated. A very common slip is to differentiate both factors in the same term — don't. Exactly one factor is differentiated per term.
乘积法则是 $f'g+fg'$,不是 $f'g'$。两项都需要,且每项让另一个因子保持不求导。一个很常见的失误是在同一项里对两个因子都求导——别这样。每项恰好对一个因子求导。
Differentiate $y=(3x^2)(\sin x)$.
- $f=3x^2\Rightarrow f'=6x$; $\quad g=\sin x\Rightarrow g'=\cos x$.
- $y'=f'g+fg'=6x\sin x+3x^2\cos x$.
- Neither factor could be simplified away, so the Product Rule was necessary.
对 $y=(3x^2)(\sin x)$ 求导。
- $f=3x^2\Rightarrow f'=6x$;$\quad g=\sin x\Rightarrow g'=\cos x$。
- $y'=f'g+fg'=6x\sin x+3x^2\cos x$。
- 两个因子都无法化简掉,所以乘积法则是必需的。
The Product Rule: $\frac{d}{dx}[fg]=f'g+fg'$ — differentiate one factor at a time and add. Use it whenever a product of differentiable functions can't be simplified first. It is not $f'g'$; each term differentiates exactly one factor.
乘积法则:$\frac{d}{dx}[fg]=f'g+fg'$——每次对一个因子求导再相加。当两个可微函数的乘积无法先化简时使用它。它不是 $f'g'$;每项恰好对一个因子求导。