Unit A3: Differentiation Rules单元 A3：求导法则

Differentiate $f(x) = 5x^3 - 7x^2 + 2x - 9$.

Apply linearity term by term, then the power rule on each power. The constant $-9$ has derivative $0$.

$$ f'(x) = 5\cdot 3x^2 - 7\cdot 2x + 2\cdot 1 - 0 = 15x^2 - 14x + 2. $$

对 $f(x) = 5x^3 - 7x^2 + 2x - 9$ 求导。

逐项使用线性性，再对每个幂使用幂法则。常数项 $-9$ 的导数为 $0$。

$$ f'(x) = 5\cdot 3x^2 - 7\cdot 2x + 2\cdot 1 - 0 = 15x^2 - 14x + 2. $$

Form the difference quotient of $f+g$ and split it using the limit laws from Unit A1.

$$ \frac{(f+g)(x+h) - (f+g)(x)}{h} = \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}. $$

Each piece converges as $h \to 0$, so the limit of the sum is the sum of the limits, giving $(f+g)'(x) = f'(x) + g'(x)$. The constant multiple rule follows the same way, factoring $c$ out of the difference quotient.

写出 $f+g$ 的差商，再用单元 A1 的极限法则（limit laws）把它拆开。

$$ \frac{(f+g)(x+h) - (f+g)(x)}{h} = \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}. $$

当 $h \to 0$ 时每一部分都收敛，因此和的极限等于极限之和，得到 $(f+g)'(x) = f'(x) + g'(x)$。常数倍法则同理可证，只需把 $c$ 从差商里提出来即可。

Differentiate $g(x) = 4\sqrt{x} + \dfrac{3}{x^2} - \dfrac{x}{2} + 6$.

Rewrite every term as a constant times a power of $x$ before differentiating: $4\sqrt{x} = 4x^{1/2}$, $\dfrac{3}{x^2} = 3x^{-2}$, and $\dfrac{x}{2} = \tfrac12 x$. The constant $6$ contributes nothing.

$$ g'(x) = 4\cdot\tfrac12 x^{-1/2} + 3\cdot(-2)x^{-3} - \tfrac12 = 2x^{-1/2} - 6x^{-3} - \tfrac12. $$

Returning to root and reciprocal notation, $g'(x) = \dfrac{2}{\sqrt{x}} - \dfrac{6}{x^3} - \dfrac12$. The rewrite step is what makes the power rule legal here, since differentiating $\sqrt{x}$ or $1/x^2$ directly is where most slips occur.

对 $g(x) = 4\sqrt{x} + \dfrac{3}{x^2} - \dfrac{x}{2} + 6$ 求导。

求导前先把每一项改写成常数乘以 $x$ 的幂：$4\sqrt{x} = 4x^{1/2}$，$\dfrac{3}{x^2} = 3x^{-2}$，$\dfrac{x}{2} = \tfrac12 x$。常数 $6$ 不贡献任何东西。

$$ g'(x) = 4\cdot\tfrac12 x^{-1/2} + 3\cdot(-2)x^{-3} - \tfrac12 = 2x^{-1/2} - 6x^{-3} - \tfrac12. $$

再换回根号与倒数记法，$g'(x) = \dfrac{2}{\sqrt{x}} - \dfrac{6}{x^3} - \dfrac12$。正是这一步改写让幂法则在这里合法可用，因为直接对 $\sqrt{x}$ 或 $1/x^2$ 求导正是大多数人出错之处。

A curve is $y = ax^4 - 5x + a^2$, where $a$ is a fixed constant. Find $y'$ and the slope at $x = 1$.

Treat $a$ and $a^2$ as constants. Only $x$ is the variable, so $a^2$ differentiates to $0$ just like any number.

$$ y' = 4ax^3 - 5. $$

At $x = 1$ the slope is $y'(1) = 4a - 5$. Recognizing which symbols are constants is half the battle in later units, where parameters appear constantly alongside the variable.

曲线为 $y = ax^4 - 5x + a^2$，其中 $a$ 是固定常数。求 $y'$ 以及在 $x = 1$ 处的斜率。

把 $a$ 与 $a^2$ 都当作常数。只有 $x$ 是变量，所以 $a^2$ 和任何数字一样，求导后为 $0$。

$$ y' = 4ax^3 - 5. $$

在 $x = 1$ 处斜率为 $y'(1) = 4a - 5$。辨认哪些符号是常数，在后续单元里已经成功了一半，因为那时参数会和变量一起频繁出现。

Find the equation of the tangent line to $f(x) = x^3 - 4x$ at $x = 1$.

Differentiate by linearity and the power rule: $f'(x) = 3x^2 - 4$. The slope at $x = 1$ is $f'(1) = 3 - 4 = -1$, and the point is $\big(1, f(1)\big) = (1, -3)$. The point-slope form gives

$$ y - (-3) = -1(x - 1) \quad\Longrightarrow\quad y = -x - 2. $$

This is the central use of the derivative: it converts an algebraic formula into the slope of the curve at a chosen point, and the differentiation rules of this unit are what make that slope cheap to compute.

求 $f(x) = x^3 - 4x$ 在 $x = 1$ 处的切线（tangent line）方程。

用线性性与幂法则求导：$f'(x) = 3x^2 - 4$。在 $x = 1$ 处的斜率为 $f'(1) = 3 - 4 = -1$，切点为 $\big(1, f(1)\big) = (1, -3)$。点斜式给出

$$ y - (-3) = -1(x - 1) \quad\Longrightarrow\quad y = -x - 2. $$

这正是导数的核心用途：它把一个代数公式转化为曲线在指定点处的斜率，而本单元的求导法则正是让这个斜率变得廉价可算的工具。

Differentiate $y = (x^2 + 1)(3x - 5)$.

Let $u = x^2 + 1$ so $u' = 2x$, and $v = 3x - 5$ so $v' = 3$. Then

$$ y' = u'v + uv' = (2x)(3x-5) + (x^2+1)(3) = 6x^2 - 10x + 3x^2 + 3 = 9x^2 - 10x + 3. $$

Expanding first and differentiating gives the same answer, a good check.

对 $y = (x^2 + 1)(3x - 5)$ 求导。

令 $u = x^2 + 1$，则 $u' = 2x$；令 $v = 3x - 5$，则 $v' = 3$。于是

$$ y' = u'v + uv' = (2x)(3x-5) + (x^2+1)(3) = 6x^2 - 10x + 3x^2 + 3 = 9x^2 - 10x + 3. $$

先展开再求导会得到相同答案，是一个不错的验算。

Add and subtract the cross term $u(x+h)\,v(x)$ in the numerator of the difference quotient.

$$ \frac{u(x+h)v(x+h) - u(x)v(x)}{h} = u(x+h)\,\frac{v(x+h)-v(x)}{h} + v(x)\,\frac{u(x+h)-u(x)}{h}. $$

As $h \to 0$, the factor $u(x+h) \to u(x)$ because $u$ is differentiable hence continuous, and the two difference quotients tend to $v'(x)$ and $u'(x)$. The limit is $u(x)v'(x) + v(x)u'(x)$.

在差商的分子里同时加上和减去交叉项 $u(x+h)\,v(x)$。

$$ \frac{u(x+h)v(x+h) - u(x)v(x)}{h} = u(x+h)\,\frac{v(x+h)-v(x)}{h} + v(x)\,\frac{u(x+h)-u(x)}{h}. $$

当 $h \to 0$ 时，因子 $u(x+h) \to u(x)$（因为 $u$ 可微，故连续），两个差商分别趋于 $v'(x)$ 与 $u'(x)$。极限即为 $u(x)v'(x) + v(x)u'(x)$。

Differentiate $y = x\,(x+2)\,(x-3)$ using the three-factor product rule.

With $u = x$, $v = x+2$, $w = x-3$ and $u' = v' = w' = 1$, the rule $(uvw)' = u'vw + uv'w + uvw'$ gives

$$ y' = (x+2)(x-3) + x(x-3) + x(x+2). $$

Expanding, $y' = (x^2 - x - 6) + (x^2 - 3x) + (x^2 + 2x) = 3x^2 - 2x - 6$. Multiplying the original out to $x^3 - x^2 - 6x$ and differentiating directly gives the same $3x^2 - 2x - 6$, confirming the three-factor pattern.

用三因子乘积法则对 $y = x\,(x+2)\,(x-3)$ 求导。

取 $u = x$、$v = x+2$、$w = x-3$，且 $u' = v' = w' = 1$，法则 $(uvw)' = u'vw + uv'w + uvw'$ 给出

$$ y' = (x+2)(x-3) + x(x-3) + x(x+2). $$

展开得 $y' = (x^2 - x - 6) + (x^2 - 3x) + (x^2 + 2x) = 3x^2 - 2x - 6$。把原式乘开为 $x^3 - x^2 - 6x$ 再直接求导，同样得到 $3x^2 - 2x - 6$，验证了三因子模式。

Differentiate $y = \sqrt{x}\,(x^2 - 4)$ and simplify.

Let $u = x^{1/2}$ so $u' = \tfrac12 x^{-1/2}$, and $v = x^2 - 4$ so $v' = 2x$. Then

$$ y' = \tfrac12 x^{-1/2}(x^2 - 4) + x^{1/2}(2x) = \frac{x^2 - 4}{2\sqrt{x}} + 2x\sqrt{x}. $$

Put both terms over $2\sqrt{x}$: $y' = \dfrac{x^2 - 4 + 4x^2}{2\sqrt{x}} = \dfrac{5x^2 - 4}{2\sqrt{x}}$. Combining over a common denominator is the standard final step when one factor is a root.

对 $y = \sqrt{x}\,(x^2 - 4)$ 求导并化简。

令 $u = x^{1/2}$，则 $u' = \tfrac12 x^{-1/2}$；令 $v = x^2 - 4$，则 $v' = 2x$。于是

$$ y' = \tfrac12 x^{-1/2}(x^2 - 4) + x^{1/2}(2x) = \frac{x^2 - 4}{2\sqrt{x}} + 2x\sqrt{x}. $$

把两项通分到 $2\sqrt{x}$：$y' = \dfrac{x^2 - 4 + 4x^2}{2\sqrt{x}} = \dfrac{5x^2 - 4}{2\sqrt{x}}$。当一个因子是根号时，通分合并是标准的收尾步骤。

Suppose $u$ and $v$ are differentiable with $u(2) = 3$, $u'(2) = -1$, $v(2) = 5$, and $v'(2) = 4$. Find $(uv)'(2)$.

The product rule needs no formula for $u$ or $v$, only their values and derivatives at the point. Substitute directly:

$$ (uv)'(2) = u'(2)\,v(2) + u(2)\,v'(2) = (-1)(5) + (3)(4) = -5 + 12 = 7. $$

This is exactly the form the rule takes on exams that give a table instead of formulas, and it underlines that the product rule is about the functions and their slopes at a point, not about any particular algebraic expression.

设 $u$ 与 $v$ 可微，且 $u(2) = 3$、$u'(2) = -1$、$v(2) = 5$、$v'(2) = 4$。求 $(uv)'(2)$。

乘积法则不需要 $u$ 或 $v$ 的具体公式，只需要它们在该点的值和导数。直接代入：

$$ (uv)'(2) = u'(2)\,v(2) + u(2)\,v'(2) = (-1)(5) + (3)(4) = -5 + 12 = 7. $$

这正是考试给出数值表而非公式时该法则呈现的形式，它强调乘积法则关心的是函数及其在某点的斜率，而不是某个特定的代数表达式。

Differentiate $f(x) = \dfrac{x}{x^2 + 1}$.

Take $u = x$ with $u' = 1$, and $v = x^2 + 1$ with $v' = 2x$. Then

$$ f'(x) = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}. $$

对 $f(x) = \dfrac{x}{x^2 + 1}$ 求导。

取 $u = x$，$u' = 1$；取 $v = x^2 + 1$，$v' = 2x$。于是

$$ f'(x) = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}. $$

Differentiate $f(x) = \dfrac{2x - 1}{x^2 + x}$.

Take $u = 2x - 1$ with $u' = 2$, and $v = x^2 + x$ with $v' = 2x + 1$. The quotient rule gives

$$ f'(x) = \frac{2(x^2 + x) - (2x - 1)(2x + 1)}{(x^2 + x)^2}. $$

Expand the numerator carefully: $2x^2 + 2x - (4x^2 - 1) = 2x^2 + 2x - 4x^2 + 1 = -2x^2 + 2x + 1$. So

$$ f'(x) = \frac{-2x^2 + 2x + 1}{(x^2 + x)^2}, \qquad x \neq 0,\ x \neq -1. $$

The minus sign in the quotient rule distributes over the whole product $(2x-1)(2x+1)$, which is exactly where sign errors creep in.

对 $f(x) = \dfrac{2x - 1}{x^2 + x}$ 求导。

取 $u = 2x - 1$，$u' = 2$；取 $v = x^2 + x$，$v' = 2x + 1$。商的法则给出

$$ f'(x) = \frac{2(x^2 + x) - (2x - 1)(2x + 1)}{(x^2 + x)^2}. $$

小心展开分子：$2x^2 + 2x - (4x^2 - 1) = 2x^2 + 2x - 4x^2 + 1 = -2x^2 + 2x + 1$。于是

$$ f'(x) = \frac{-2x^2 + 2x + 1}{(x^2 + x)^2}, \qquad x \neq 0,\ x \neq -1. $$

商的法则里的减号要分配到整个乘积 $(2x-1)(2x+1)$ 上，这正是符号错误最容易溜进来的地方。

Differentiate $f(x) = \dfrac{1}{x^2 + 3x + 5}$ using the reciprocal rule.

The reciprocal rule $\dfrac{d}{dx}\dfrac{1}{v} = -\dfrac{v'}{v^2}$ avoids the full quotient rule whenever the numerator is constant. Here $v = x^2 + 3x + 5$ with $v' = 2x + 3$, so

$$ f'(x) = -\frac{2x + 3}{(x^2 + 3x + 5)^2}. $$

The full quotient rule with $u = 1$, $u' = 0$ reproduces the same thing, but the reciprocal form is faster and less error-prone when the top is just a number.

用倒数法则对 $f(x) = \dfrac{1}{x^2 + 3x + 5}$ 求导。

每当分子是常数时，倒数法则 $\dfrac{d}{dx}\dfrac{1}{v} = -\dfrac{v'}{v^2}$ 可以省去完整的商的法则。这里 $v = x^2 + 3x + 5$，$v' = 2x + 3$，所以

$$ f'(x) = -\frac{2x + 3}{(x^2 + 3x + 5)^2}. $$

取 $u = 1$、$u' = 0$ 用完整的商的法则会得到同样结果，但当分子只是一个数字时，倒数形式更快、也更不容易出错。

You do not have to take the quotient rule on faith. Write $f = u/v$ as the product $f = u\cdot v^{-1}$ and differentiate with the product rule, using the chain rule on $v^{-1}$.

$$ f' = u'\,v^{-1} + u\cdot(-1)v^{-2}v' = \frac{u'}{v} - \frac{u\,v'}{v^2}. $$

Put both terms over the common denominator $v^2$:

$$ f' = \frac{u'v}{v^2} - \frac{u\,v'}{v^2} = \frac{u'v - u\,v'}{v^2}. $$

This is the quotient rule, and the derivation also explains the structure: the squared denominator comes from the $v^{-2}$ in the chain step, and the minus sign comes from the derivative of $v^{-1}$. Seeing it this way means you only ever need to memorize the product and chain rules.

你不必盲信商的法则。把 $f = u/v$ 写成乘积 $f = u\cdot v^{-1}$，用乘积法则求导，并对 $v^{-1}$ 使用链式法则（Chain Rule）。

$$ f' = u'\,v^{-1} + u\cdot(-1)v^{-2}v' = \frac{u'}{v} - \frac{u\,v'}{v^2}. $$

把两项通分到公分母 $v^2$：

$$ f' = \frac{u'v}{v^2} - \frac{u\,v'}{v^2} = \frac{u'v - u\,v'}{v^2}. $$

这就是商的法则，推导也解释了它的结构：平方的分母来自链式步骤中的 $v^{-2}$，减号来自 $v^{-1}$ 的导数。这样看，你只需要记住乘积法则和链式法则即可。

Differentiate $y = (3x^2 + 1)^4$.

The outer function is $u^4$ and the inner function is $u = 3x^2 + 1$, with $u' = 6x$. By the general power rule,

$$ y' = 4(3x^2 + 1)^3 \cdot 6x = 24x\,(3x^2 + 1)^3. $$

对 $y = (3x^2 + 1)^4$ 求导。

外层函数是 $u^4$，内层函数是 $u = 3x^2 + 1$，$u' = 6x$。由广义幂法则，

$$ y' = 4(3x^2 + 1)^3 \cdot 6x = 24x\,(3x^2 + 1)^3. $$

Differentiate $y = \sqrt{\,\sin(x^2)\,}$. Peel one layer at a time from the outside in.

$$ y' = \frac{1}{2\sqrt{\sin(x^2)}}\cdot \cos(x^2)\cdot 2x = \frac{x\cos(x^2)}{\sqrt{\sin(x^2)}}. $$

Each application of the chain rule contributes one factor: the square root layer, the sine layer, and finally the derivative of $x^2$.

对 $y = \sqrt{\,\sin(x^2)\,}$ 求导。从外往里一层一层地剥。

$$ y' = \frac{1}{2\sqrt{\sin(x^2)}}\cdot \cos(x^2)\cdot 2x = \frac{x\cos(x^2)}{\sqrt{\sin(x^2)}}. $$

链式法则每用一次就贡献一个因子：根号层、正弦层，最后是 $x^2$ 的导数。

Differentiate $y = \sqrt{4x^2 + 9} = (4x^2 + 9)^{1/2}$.

The outer function is $u^{1/2}$ with derivative $\tfrac12 u^{-1/2}$, and the inner function is $u = 4x^2 + 9$ with $u' = 8x$. The general power rule gives

$$ y' = \tfrac12 (4x^2 + 9)^{-1/2}\cdot 8x = \frac{4x}{\sqrt{4x^2 + 9}}. $$

Square roots are just the exponent $1/2$; rewriting them that way makes the chain rule entirely routine.

对 $y = \sqrt{4x^2 + 9} = (4x^2 + 9)^{1/2}$ 求导。

外层函数是 $u^{1/2}$，导数为 $\tfrac12 u^{-1/2}$；内层函数是 $u = 4x^2 + 9$，$u' = 8x$。广义幂法则给出

$$ y' = \tfrac12 (4x^2 + 9)^{-1/2}\cdot 8x = \frac{4x}{\sqrt{4x^2 + 9}}. $$

根号无非就是 $1/2$ 次幂；把它这样改写，链式法则就变得完全例行化。

Differentiate $y = \dfrac{1}{(x^3 + 1)^2}$ by writing it as a negative power.

Rewrite $y = (x^3 + 1)^{-2}$. The outer function is $u^{-2}$ with derivative $-2u^{-3}$, and the inner is $u = x^3 + 1$ with $u' = 3x^2$. The general power rule gives

$$ y' = -2(x^3 + 1)^{-3}\cdot 3x^2 = -\frac{6x^2}{(x^3 + 1)^3}. $$

Recasting a reciprocal as a negative power often turns a quotient-rule problem into a one-line chain-rule problem. Recognizing when to rewrite is a skill worth practicing deliberately.

把 $y = \dfrac{1}{(x^3 + 1)^2}$ 写成负指数幂再求导。

改写 $y = (x^3 + 1)^{-2}$。外层函数是 $u^{-2}$，导数为 $-2u^{-3}$；内层是 $u = x^3 + 1$，$u' = 3x^2$。广义幂法则给出

$$ y' = -2(x^3 + 1)^{-3}\cdot 3x^2 = -\frac{6x^2}{(x^3 + 1)^3}. $$

把倒数改写成负指数幂，常常能把一道商的法则题变成一行就能解决的链式法则题。识别何时该改写，是一项值得刻意练习的技能。

Set $y = f(g(x))$ and $u = g(x)$. The naive argument multiplies and divides by $\Delta u = g(x+h) - g(x)$:

$$ \frac{f(g(x+h)) - f(g(x))}{h} = \frac{f(u + \Delta u) - f(u)}{\Delta u}\cdot\frac{\Delta u}{h}. $$

As $h \to 0$, continuity of $g$ forces $\Delta u \to 0$, so the first factor tends to $f'(u)$ and the second to $g'(x)$, giving $f'(g(x))\,g'(x)$.

The subtlety is that $\Delta u$ can equal $0$ for $h$ near $0$, making the division illegal. The honest fix defines an auxiliary function $E(\Delta u)$ equal to $\dfrac{f(u+\Delta u) - f(u)}{\Delta u} - f'(u)$ when $\Delta u \neq 0$ and equal to $0$ when $\Delta u = 0$. Then $E$ is continuous at $0$ with $E(0) = 0$, and $f(u+\Delta u) - f(u) = \big(f'(u) + E(\Delta u)\big)\Delta u$ holds for every $\Delta u$. Dividing by $h$ and letting $h \to 0$ gives $f'(u)\,g'(x)$ cleanly. This is the standard rigorous proof and the reason careful texts introduce that error term.

令 $y = f(g(x))$，$u = g(x)$。朴素的论证是同乘同除 $\Delta u = g(x+h) - g(x)$：

$$ \frac{f(g(x+h)) - f(g(x))}{h} = \frac{f(u + \Delta u) - f(u)}{\Delta u}\cdot\frac{\Delta u}{h}. $$

当 $h \to 0$ 时，$g$ 的连续性迫使 $\Delta u \to 0$，于是第一个因子趋于 $f'(u)$，第二个趋于 $g'(x)$，得到 $f'(g(x))\,g'(x)$。

微妙之处在于：当 $h$ 接近 $0$ 时 $\Delta u$ 可能等于 $0$，使得该除法非法。严谨的补救是定义一个辅助函数 $E(\Delta u)$：当 $\Delta u \neq 0$ 时它等于 $\dfrac{f(u+\Delta u) - f(u)}{\Delta u} - f'(u)$，当 $\Delta u = 0$ 时它等于 $0$。这样 $E$ 在 $0$ 处连续且 $E(0) = 0$，并且 $f(u+\Delta u) - f(u) = \big(f'(u) + E(\Delta u)\big)\Delta u$ 对每个 $\Delta u$ 都成立。除以 $h$ 并令 $h \to 0$，干净地得到 $f'(u)\,g'(x)$。这是标准的严格证明，也是严谨教材引入那个误差项的原因。

Differentiate $y = x^2 (2x + 1)^3$.

The last operation is multiplication, so use the product rule with $u = x^2$ and $v = (2x+1)^3$. The factor $v$ needs the chain rule: $v' = 3(2x+1)^2\cdot 2 = 6(2x+1)^2$.

$$ y' = 2x\,(2x+1)^3 + x^2\cdot 6(2x+1)^2 = 2x(2x+1)^2\big[(2x+1) + 3x\big] = 2x(2x+1)^2(5x + 1). $$

对 $y = x^2 (2x + 1)^3$ 求导。

最后一个运算是乘法，所以用乘积法则，取 $u = x^2$、$v = (2x+1)^3$。因子 $v$ 需要链式法则：$v' = 3(2x+1)^2\cdot 2 = 6(2x+1)^2$。

$$ y' = 2x\,(2x+1)^3 + x^2\cdot 6(2x+1)^2 = 2x(2x+1)^2\big[(2x+1) + 3x\big] = 2x(2x+1)^2(5x + 1). $$

Differentiate $y = \dfrac{x(x+1)}{x - 2}$.

The last operation is division, so the quotient rule is outermost with $u = x(x+1) = x^2 + x$ and $v = x - 2$. Here $u' = 2x + 1$ after expanding, so no nested product rule is needed, and $v' = 1$.

$$ y' = \frac{(2x+1)(x-2) - (x^2 + x)(1)}{(x-2)^2}. $$

Expand the numerator: $(2x^2 - 3x - 2) - (x^2 + x) = x^2 - 4x - 2$. Therefore

$$ y' = \frac{x^2 - 4x - 2}{(x-2)^2}, \qquad x \neq 2. $$

Expanding $u$ first sidestepped a nested product rule. Choosing whether to expand or to keep factors is a judgment call: expand when the algebra is short, keep factored form when it is not.

对 $y = \dfrac{x(x+1)}{x - 2}$ 求导。

最后一个运算是除法，所以商的法则在最外层，取 $u = x(x+1) = x^2 + x$、$v = x - 2$。这里展开后 $u' = 2x + 1$，因此不需要嵌套的乘积法则，且 $v' = 1$。

$$ y' = \frac{(2x+1)(x-2) - (x^2 + x)(1)}{(x-2)^2}. $$

展开分子：$(2x^2 - 3x - 2) - (x^2 + x) = x^2 - 4x - 2$。因此

$$ y' = \frac{x^2 - 4x - 2}{(x-2)^2}, \qquad x \neq 2. $$

先展开 $u$ 就绕过了一次嵌套的乘积法则。选择展开还是保留因式是一种判断：代数运算短就展开，否则就保留因式形式。

Differentiate $y = \big(x^2 + (2x+1)^3\big)^5$.

Outermost is the fifth power, so $y' = 5\big(x^2 + (2x+1)^3\big)^4\cdot\dfrac{d}{dx}\big(x^2 + (2x+1)^3\big)$. The inside is a sum; its derivative uses linearity and a second chain rule:

$$ \frac{d}{dx}\big(x^2 + (2x+1)^3\big) = 2x + 3(2x+1)^2\cdot 2 = 2x + 6(2x+1)^2. $$

Assembling,

$$ y' = 5\big(x^2 + (2x+1)^3\big)^4\big(2x + 6(2x+1)^2\big). $$

Each layer was handled by exactly one rule, applied from the outside in. The discipline of writing the unfinished derivative with a placeholder before filling the inside keeps long chains organized.

对 $y = \big(x^2 + (2x+1)^3\big)^5$ 求导。

最外层是五次幂，所以 $y' = 5\big(x^2 + (2x+1)^3\big)^4\cdot\dfrac{d}{dx}\big(x^2 + (2x+1)^3\big)$。内层是一个和；它的导数用线性性和第二次链式法则：

$$ \frac{d}{dx}\big(x^2 + (2x+1)^3\big) = 2x + 3(2x+1)^2\cdot 2 = 2x + 6(2x+1)^2. $$

组装起来，

$$ y' = 5\big(x^2 + (2x+1)^3\big)^4\big(2x + 6(2x+1)^2\big). $$

每一层都恰好由一条法则处理，从外往里依次应用。在填内层之前先用占位符写下未完成的导数，这一纪律能让长链条保持条理。

Differentiate $R(x) = \dfrac{x^2 - 1}{x + 2}$ and note where it is valid.

With $P = x^2 - 1$ ($P' = 2x$) and $Q = x + 2$ ($Q' = 1$),

$$ R'(x) = \frac{2x(x+2) - (x^2 - 1)(1)}{(x+2)^2} = \frac{2x^2 + 4x - x^2 + 1}{(x+2)^2} = \frac{x^2 + 4x + 1}{(x+2)^2}, \quad x \neq -2. $$

对 $R(x) = \dfrac{x^2 - 1}{x + 2}$ 求导，并注明它在何处有效。

取 $P = x^2 - 1$（$P' = 2x$）、$Q = x + 2$（$Q' = 1$），

$$ R'(x) = \frac{2x(x+2) - (x^2 - 1)(1)}{(x+2)^2} = \frac{2x^2 + 4x - x^2 + 1}{(x+2)^2} = \frac{x^2 + 4x + 1}{(x+2)^2}, \quad x \neq -2. $$

For $p(x) = 2x^4 - x^3 + 5x$, find the first three derivatives.

Differentiate term by term each time:

$$ p'(x) = 8x^3 - 3x^2 + 5, \qquad p''(x) = 24x^2 - 6x, \qquad p'''(x) = 48x - 6. $$

The degree drops by one at every step, from $4$ to $3$ to $2$ to $1$. One more derivative gives the constant $p^{(4)}(x) = 48$, and the next is $p^{(5)}(x) = 0$.

对 $p(x) = 2x^4 - x^3 + 5x$，求前三阶导数。

每次都逐项求导：

$$ p'(x) = 8x^3 - 3x^2 + 5, \qquad p''(x) = 24x^2 - 6x, \qquad p'''(x) = 48x - 6. $$

次数每步降一，从 $4$ 到 $3$ 到 $2$ 到 $1$。再求一次导得常数 $p^{(4)}(x) = 48$，下一次则是 $p^{(5)}(x) = 0$。

Differentiate $R(x) = \dfrac{x^2}{x - 1}$ and locate where the slope is zero.

With $P = x^2$ so $P' = 2x$, and $Q = x - 1$ so $Q' = 1$,

$$ R'(x) = \frac{2x(x-1) - x^2(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}. $$

The numerator vanishes at $x = 0$ and $x = 2$, so the tangent is horizontal there. The point $x = 1$ is a genuine vertical asymptote, not a zero of $R'$. Factoring the numerator is what exposes the critical points cleanly.

对 $R(x) = \dfrac{x^2}{x - 1}$ 求导，并找出斜率为零的位置。

取 $P = x^2$（$P' = 2x$）、$Q = x - 1$（$Q' = 1$），

$$ R'(x) = \frac{2x(x-1) - x^2(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}. $$

分子在 $x = 0$ 和 $x = 2$ 处为零，所以切线在那里水平。点 $x = 1$ 是真正的竖直渐近线，而不是 $R'$ 的零点。把分子因式分解，正是干净地暴露临界点的方法。

A student writes $\dfrac{d}{dx}\sin(5x) = \cos(5x)$. What is missing?

The argument $5x$ is an inner function with derivative $5$, so the chain rule requires

$$ \frac{d}{dx}\sin(5x) = \cos(5x)\cdot 5 = 5\cos(5x). $$

The corrected answer carries the factor $5$ that the inner derivative supplies.

一名学生写道 $\dfrac{d}{dx}\sin(5x) = \cos(5x)$。漏了什么？

自变量 $5x$ 是一个内层函数，其导数为 $5$，所以链式法则要求

$$ \frac{d}{dx}\sin(5x) = \cos(5x)\cdot 5 = 5\cos(5x). $$

修正后的答案带上了内层导数提供的因子 $5$。

A student claims $\dfrac{d}{dx}\dfrac{x^2}{x+1} = \dfrac{2x}{1} = 2x$. Find and fix the error.

The student differentiated numerator and denominator separately, the false rule $\left(\tfrac{u}{v}\right)' = \tfrac{u'}{v'}$. The correct quotient rule with $u = x^2$ and $v = x+1$ gives

$$ \frac{d}{dx}\frac{x^2}{x+1} = \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}. $$

The two answers disagree even at a single test point: at $x = 1$ the wrong rule gives $2$, while the correct derivative gives $3/4$. Plugging in one number is a fast way to catch a structural error before it propagates.

一名学生声称 $\dfrac{d}{dx}\dfrac{x^2}{x+1} = \dfrac{2x}{1} = 2x$。找出并改正错误。

该学生把分子和分母分别求导，用了错误法则 $\left(\tfrac{u}{v}\right)' = \tfrac{u'}{v'}$。取 $u = x^2$、$v = x+1$，正确的商的法则给出

$$ \frac{d}{dx}\frac{x^2}{x+1} = \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}. $$

两个答案即使在单个测试点也不一致：在 $x = 1$ 处错误法则给出 $2$，而正确导数给出 $3/4$。代入一个数字，是在结构性错误扩散之前快速抓住它的好办法。

Differentiate $y = \dfrac{(x^2 + 1)^3}{x - 4}$, which combines the quotient, chain, and power rules.

Outermost is division, so the quotient rule applies with $u = (x^2+1)^3$ and $v = x - 4$. The numerator derivative needs the chain rule: $u' = 3(x^2+1)^2\cdot 2x = 6x(x^2+1)^2$, and $v' = 1$. Then

$$ y' = \frac{6x(x^2+1)^2(x-4) - (x^2+1)^3(1)}{(x-4)^2}. $$

Factor $(x^2+1)^2$ from the numerator:

$$ y' = \frac{(x^2+1)^2\big[6x(x-4) - (x^2+1)\big]}{(x-4)^2} = \frac{(x^2+1)^2(5x^2 - 24x - 1)}{(x-4)^2}, \quad x \neq 4. $$

Three rules, applied outside in, with a common factor pulled out at the end. This is the template for every hard derivative in the next several units.

对 $y = \dfrac{(x^2 + 1)^3}{x - 4}$ 求导，它综合了商的法则、链式法则与幂法则。

最外层是除法，所以用商的法则，取 $u = (x^2+1)^3$、$v = x - 4$。分子的导数需要链式法则：$u' = 3(x^2+1)^2\cdot 2x = 6x(x^2+1)^2$，且 $v' = 1$。于是

$$ y' = \frac{6x(x^2+1)^2(x-4) - (x^2+1)^3(1)}{(x-4)^2}. $$

从分子提出公因子 $(x^2+1)^2$：

$$ y' = \frac{(x^2+1)^2\big[6x(x-4) - (x^2+1)\big]}{(x-4)^2} = \frac{(x^2+1)^2(5x^2 - 24x - 1)}{(x-4)^2}, \quad x \neq 4. $$

三条法则，从外往里应用，最后提出一个公因子。这就是接下来几个单元里每一道难求导题的模板。