Unit E2: Techniques
of Differential Calculus单元 E2：微分技巧
The toolkit unit. Where E1 builds the derivative from a limit, E2 supplies the four rules that let you differentiate every elementary function in the AA syllabus: the chain rule for composites, the product and quotient rules for combinations, and the standard derivatives of $\sin x$, $\cos x$, $\tan x$, $e^{x}$, and $\ln x$. The HL sections add implicit differentiation (for curves where $y$ cannot be isolated) and logarithmic differentiation (for awkward products of powers).本单元是工具箱。E1 由极限建立导数，E2 给出可对 AA 大纲中每个初等函数求导的四条规则：复合函数的链式法则（chain rule），组合函数的乘积与商法则（product / quotient rules），以及 $\sin x$、$\cos x$、$\tan x$、$e^{x}$、$\ln x$ 的标准导数。HL 部分增补隐函数求导（适用于 $y$ 无法显式解出的曲线）和对数求导法（适用于幂次组合复杂的函数）。
How to use this guide本指南使用说明
E2 is computational. There are six rules; learn them once, drill them on every assigned function, and they become reflexive. Almost every Paper 1 differentiation question reduces to identifying which rule to apply first.E2 偏计算。共六条规则，记一次、对每个函数练熟，就能形成条件反射。Paper 1 几乎所有求导题都可归结为"先判断该用哪条规则"。
Memorise the six standard derivatives ($\sin$, $\cos$, $\tan$, $e^{x}$, $\ln x$, $x^{n}$) plus the chain, product, and quotient rule statements. Skip implicit and logarithmic differentiation if you are out of time; they are HL only. Run three practice problems per rule.
背熟六个标准导数（$\sin$、$\cos$、$\tan$、$e^{x}$、$\ln x$、$x^{n}$）以及链式、乘积、商法则的公式。时间不够可跳过隐函数与对数求导（HL 专属）。每条规则做三道题。
链式法则要会"嵌套两层及以上"的情形（如 $\sin(\ln(x^{2} + 1))$）。隐函数求导后必须解出 $\tfrac{dy}{dx}$ 的封闭式。对数求导要会处理 $y = (x+1)^{x}$ 这类"变底变指"组合。请背 $\tfrac{d}{dx}\arcsin x = 1/\sqrt{1 - x^{2}}$ 等反三角导数。
Chain rule must handle two or more nested layers (e.g. $\sin(\ln(x^{2} + 1))$). After implicit differentiation, isolate $\tfrac{dy}{dx}$ as a closed-form expression. Logarithmic differentiation must handle $y = (x+1)^{x}$ style "variable base, variable exponent" combinations. Memorise the inverse trig derivatives ($\tfrac{d}{dx}\arcsin x = 1/\sqrt{1 - x^{2}}$, etc.).
Power Rule and Linearity幂法则与线性性 SL 5.3
- Power rule. For any real $n$, $\dfrac{d}{dx}(x^{n}) = n x^{n-1}$.
- Constant multiple. $\dfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$.
- Sum and difference. $\dfrac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$.
- 幂法则。对任意实数 $n$，$\dfrac{d}{dx}(x^{n}) = n x^{n-1}$。
- 常数倍法则。$\dfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$。
- 和差法则。$\dfrac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$。
Differentiate $h(x) = 5x^{4} - \dfrac{3}{x^{2}} + 7\sqrt{x} - 11$.求 $h(x) = 5x^{4} - \dfrac{3}{x^{2}} + 7\sqrt{x} - 11$ 的导数。
Step 1. Rewrite every term as a power of $x$.
第 1 步。将每一项化为 $x$ 的幂。
$$ h(x) \;=\; 5 x^{4} \;-\; 3 x^{-2} \;+\; 7 x^{1/2} \;-\; 11. $$Step 2. Apply the power rule term by term.
第 2 步。逐项使用幂法则。
$$ h'(x) \;=\; 20 x^{3} \;+\; 6 x^{-3} \;+\; \tfrac{7}{2} x^{-1/2} \;-\; 0. $$Step 3. Rewrite in the requested form.
第 3 步。按题目要求改写。
$$ h'(x) \;=\; 20 x^{3} \;+\; \dfrac{6}{x^{3}} \;+\; \dfrac{7}{2\sqrt{x}}. $$The Chain Rule链式法则 SL 5.7
Differentiate $y = (3 x^{2} + 1)^{5}$.求 $y = (3 x^{2} + 1)^{5}$ 的导数。
Identify outer and inner. Outer: $f(u) = u^{5}$. Inner: $u = g(x) = 3 x^{2} + 1$.
识别外层与内层。外层：$f(u) = u^{5}$。内层：$u = g(x) = 3 x^{2} + 1$。
Compute each piece. $f'(u) = 5 u^{4}$ and $g'(x) = 6x$.
分别求导。$f'(u) = 5 u^{4}$，$g'(x) = 6x$。
Combine.
相乘。
$$ \frac{dy}{dx} \;=\; 5 (3 x^{2} + 1)^{4} \cdot 6x \;=\; 30 x (3 x^{2} + 1)^{4}. $$Differentiate $y = \sin\!\bigl( \ln(x^{2} + 1) \bigr)$.求 $y = \sin\!\bigl( \ln(x^{2} + 1) \bigr)$ 的导数。
Identify three layers. Outer: $\sin(\cdot)$. Middle: $\ln(\cdot)$. Inner: $x^{2} + 1$.
识别三层。外层：$\sin(\cdot)$。中层：$\ln(\cdot)$。内层：$x^{2} + 1$。
Differentiate each layer, evaluated at the next layer in.
逐层求导，每层以"下一层"为自变量。
- Outer: $\dfrac{d}{du} \sin u = \cos u$, evaluated at $u = \ln(x^{2} + 1)$.
- 外层：$\dfrac{d}{du} \sin u = \cos u$，代入 $u = \ln(x^{2} + 1)$。
- Middle: $\dfrac{d}{dv} \ln v = \dfrac{1}{v}$, evaluated at $v = x^{2} + 1$.
- 中层：$\dfrac{d}{dv} \ln v = \dfrac{1}{v}$，代入 $v = x^{2} + 1$。
- Inner: $\dfrac{d}{dx}(x^{2} + 1) = 2x$.
- 内层：$\dfrac{d}{dx}(x^{2} + 1) = 2x$。
Multiply.
三者相乘。
$$ \frac{dy}{dx} \;=\; \cos\!\bigl(\ln(x^{2} + 1)\bigr) \cdot \frac{1}{x^{2} + 1} \cdot 2x \;=\; \frac{2x \,\cos\!\bigl(\ln(x^{2} + 1)\bigr)}{x^{2} + 1}. $$▸ Going deeper: why the chain rule has this shape▸ 深入：链式法则为何呈此形态
A short proof using the first-principles definition (sketch). For small $\Delta x$, define $\Delta u = g(x + \Delta x) - g(x)$ and $\Delta y = f(u + \Delta u) - f(u)$. Then
用 first principles 给出粗略证明。对小的 $\Delta x$，设 $\Delta u = g(x + \Delta x) - g(x)$ 与 $\Delta y = f(u + \Delta u) - f(u)$。则
$$ \frac{\Delta y}{\Delta x} \;=\; \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}, $$provided $\Delta u \ne 0$. Taking $\Delta x \to 0$ forces $\Delta u \to 0$ (by continuity of $g$), so the right side tends to $\tfrac{dy}{du} \cdot \tfrac{du}{dx}$. A rigorous proof handles the $\Delta u = 0$ case separately, but the algebraic identity is the heart of it. This is why the chain rule reads as a product of derivatives.
前提是 $\Delta u \ne 0$。令 $\Delta x \to 0$ 则 $\Delta u \to 0$（由 $g$ 的连续性），故右侧趋于 $\tfrac{dy}{du} \cdot \tfrac{du}{dx}$。严格证明须单独处理 $\Delta u = 0$，但代数恒等式就是该法则的核心。这就是链式法则呈"导数之积"形态的原因。
Product Rule and Quotient Rule乘积法则与商法则 SL 5.7
Differentiate $y = (x^{2} + 1)(x^{3} - 2x)$.求 $y = (x^{2} + 1)(x^{3} - 2x)$ 的导数。
Label. $u(x) = x^{2} + 1$, $v(x) = x^{3} - 2x$. Then $u'(x) = 2x$ and $v'(x) = 3 x^{2} - 2$.
记号。$u(x) = x^{2} + 1$，$v(x) = x^{3} - 2x$。$u'(x) = 2x$，$v'(x) = 3 x^{2} - 2$。
Apply the product rule.
使用乘积法则。
$$ \frac{dy}{dx} \;=\; 2x \,(x^{3} - 2x) \;+\; (x^{2} + 1)(3 x^{2} - 2). $$Expand and collect.
展开合并。
$$ \frac{dy}{dx} \;=\; 2 x^{4} - 4 x^{2} + 3 x^{4} - 2 x^{2} + 3 x^{2} - 2 \;=\; 5 x^{4} - 3 x^{2} - 2. $$Sanity check. Expand $y$ first: $y = x^{5} - 2 x^{3} + x^{3} - 2x = x^{5} - x^{3} - 2x$. Power rule gives $y' = 5 x^{4} - 3 x^{2} - 2$. The two answers agree.
验证。先展开 $y$：$y = x^{5} - 2 x^{3} + x^{3} - 2x = x^{5} - x^{3} - 2x$。幂法则给出 $y' = 5 x^{4} - 3 x^{2} - 2$。两种方法结果一致。
Differentiate $y = \dfrac{x}{x^{2} + 1}$.求 $y = \dfrac{x}{x^{2} + 1}$ 的导数。
Label. $u(x) = x$ (so $u'(x) = 1$); $v(x) = x^{2} + 1$ (so $v'(x) = 2x$).
记号。$u(x) = x$（$u'(x) = 1$）；$v(x) = x^{2} + 1$（$v'(x) = 2x$）。
Apply the quotient rule.
使用商法则。
$$ \frac{dy}{dx} \;=\; \frac{1 \cdot (x^{2} + 1) \;-\; x \cdot 2x}{(x^{2} + 1)^{2}} \;=\; \frac{x^{2} + 1 \;-\; 2 x^{2}}{(x^{2} + 1)^{2}} \;=\; \frac{1 - x^{2}}{(x^{2} + 1)^{2}}. $$▸ Going deeper: deriving the quotient rule from the product rule▸ 深入：由乘积法则推商法则
Write $y = u/v$ as $y \cdot v = u$. Differentiate both sides using the product rule on the left:
把 $y = u/v$ 改写为 $y \cdot v = u$。两边对 $x$ 求导，左侧用乘积法则：
$$ y' v + y v' \;=\; u' \;\Longrightarrow\; y' v \;=\; u' - y v' \;=\; u' - \tfrac{u}{v} v'. $$Divide by $v$ and clear fractions:
两边除以 $v$ 并通分：
$$ y' \;=\; \frac{u' v - u v'}{v^{2}}. $$So the quotient rule is not an independent axiom; it is a consequence of the product rule plus implicit differentiation. Memorising the product rule alone is enough if you can do this derivation under exam time pressure.
商法则并非独立公理，而是乘积法则加隐函数求导的推论。若考试时能临场推导，只背乘积法则也够用。
Derivatives of Trigonometric, Exponential, and Logarithmic Functions三角、指数、对数函数的导数 SL 5.7
| Function | Derivative |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^{2} x$ |
| $e^{x}$ | $e^{x}$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $a^{x}$ | $a^{x} \ln a$ |
| $\log_{a} x$ | $\dfrac{1}{x \ln a}$ |
Combine with the chain rule whenever the argument is more complex than $x$ alone: $\dfrac{d}{dx} \sin(g(x)) = \cos(g(x)) \cdot g'(x)$, and so on.
| 函数 | 导数 |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^{2} x$ |
| $e^{x}$ | $e^{x}$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $a^{x}$ | $a^{x} \ln a$ |
| $\log_{a} x$ | $\dfrac{1}{x \ln a}$ |
当自变量比单独的 $x$ 更复杂时，与链式法则结合：$\dfrac{d}{dx} \sin(g(x)) = \cos(g(x)) \cdot g'(x)$，其余类推。
Differentiate $y = e^{\sin(2x)}$.求 $y = e^{\sin(2x)}$ 的导数。
Three layers. Outer $e^{(\cdot)}$, middle $\sin(\cdot)$, inner $2x$.
三层结构。外层 $e^{(\cdot)}$，中层 $\sin(\cdot)$，内层 $2x$。
Apply the chain rule across all three.
对三层依次使用链式法则。
$$ \frac{dy}{dx} \;=\; e^{\sin(2x)} \cdot \cos(2x) \cdot 2 \;=\; 2 \cos(2x) \, e^{\sin(2x)}. $$Differentiate $y = x^{2} \ln x$.求 $y = x^{2} \ln x$ 的导数。
Product rule with $u = x^{2}$, $v = \ln x$:
乘积法则，$u = x^{2}$，$v = \ln x$：
$$ \frac{dy}{dx} \;=\; 2x \cdot \ln x \;+\; x^{2} \cdot \frac{1}{x} \;=\; 2x \ln x \;+\; x \;=\; x \,(2 \ln x + 1). $$▸ Going deeper: where $\tfrac{d}{dx}(e^{x}) = e^{x}$ comes from▸ 深入：$\tfrac{d}{dx}(e^{x}) = e^{x}$ 的由来
From first principles,
由定义出发，
$$ \frac{d}{dx}(e^{x}) \;=\; \lim_{h \to 0} \frac{e^{x + h} - e^{x}}{h} \;=\; e^{x} \cdot \lim_{h \to 0} \frac{e^{h} - 1}{h}. $$The number $e$ is defined precisely so that $\lim_{h \to 0} \tfrac{e^{h} - 1}{h} = 1$. (Equivalently, $e$ is the unique base for which the slope of $a^{x}$ at $x = 0$ equals $1$.) Hence $\tfrac{d}{dx}(e^{x}) = e^{x} \cdot 1 = e^{x}$. The same calculation gives $\tfrac{d}{dx}(a^{x}) = a^{x} \ln a$, since $\lim_{h \to 0} \tfrac{a^{h} - 1}{h} = \ln a$ in general.
数 $e$ 的定义就是使 $\lim_{h \to 0} \tfrac{e^{h} - 1}{h} = 1$ 的那个底数（等价地：使 $a^{x}$ 在 $x = 0$ 处斜率为 $1$ 的唯一底）。故 $\tfrac{d}{dx}(e^{x}) = e^{x} \cdot 1 = e^{x}$。同理可得 $\tfrac{d}{dx}(a^{x}) = a^{x} \ln a$，因为一般来说 $\lim_{h \to 0} \tfrac{a^{h} - 1}{h} = \ln a$。
Implicit Differentiation隐函数求导 HL AHL 5.13
The technique (three steps).
- Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.
- Every $y$-term gets a chain-rule factor of $\dfrac{dy}{dx}$. So $\dfrac{d}{dx}(y^{2}) = 2y \, \dfrac{dy}{dx}$, and $\dfrac{d}{dx}(xy) = y + x \dfrac{dy}{dx}$ (product rule).
- Collect all $\dfrac{dy}{dx}$ terms on one side and solve algebraically.
implicit differentiation）完全绕开"选分支"。
三步法。
- 两边同时对 $x$ 求导，把 $y$ 当作 $x$ 的函数。
- 每个含 $y$ 的项都要多出一个链式因子 $\dfrac{dy}{dx}$。如 $\dfrac{d}{dx}(y^{2}) = 2y \, \dfrac{dy}{dx}$；$\dfrac{d}{dx}(xy) = y + x \dfrac{dy}{dx}$（乘积法则）。
- 把所有含 $\dfrac{dy}{dx}$ 的项移到一侧，代数解出 $\dfrac{dy}{dx}$。
Find $\dfrac{dy}{dx}$ for the circle $x^{2} + y^{2} = 25$, and use it to find the slope of the tangent at the point $(3, 4)$.求圆 $x^{2} + y^{2} = 25$ 的 $\dfrac{dy}{dx}$，并由此求 $(3, 4)$ 处切线的斜率。
Differentiate both sides.
两边求导。
$$ 2x \;+\; 2y \,\frac{dy}{dx} \;=\; 0. $$Solve for $\dfrac{dy}{dx}$.
解出 $\dfrac{dy}{dx}$。
$$ \frac{dy}{dx} \;=\; -\frac{x}{y}. $$Substitute the point. At $(3, 4)$: $\dfrac{dy}{dx} = -\dfrac{3}{4}$.
代入点。$(3, 4)$ 处：$\dfrac{dy}{dx} = -\dfrac{3}{4}$。
Geometric check. The radius from the origin to $(3, 4)$ has slope $4/3$. The tangent is perpendicular to that radius, so its slope is $-3/4$. The implicit derivative agrees.
几何验证。原点到 $(3, 4)$ 的半径斜率为 $4/3$。切线与半径垂直，斜率应为 $-3/4$。隐函数导数结果一致。
Find $\dfrac{dy}{dx}$ for $x^{3} + 3 x y + y^{3} = 1$.求 $x^{3} + 3 x y + y^{3} = 1$ 的 $\dfrac{dy}{dx}$。
Differentiate term by term. The middle term $3xy$ needs the product rule.
逐项求导。中间项 $3xy$ 须用乘积法则。
$$ 3 x^{2} \;+\; 3\!\left[y + x \,\frac{dy}{dx}\right] \;+\; 3 y^{2} \,\frac{dy}{dx} \;=\; 0. $$Distribute and collect.
展开并合并同类项。
$$ 3 x^{2} + 3 y \;+\; \bigl(3x + 3 y^{2}\bigr) \frac{dy}{dx} \;=\; 0. $$Solve.
求解。
$$ \frac{dy}{dx} \;=\; -\frac{3 x^{2} + 3 y}{3 x + 3 y^{2}} \;=\; -\frac{x^{2} + y}{x + y^{2}}. $$Logarithmic Differentiation对数求导法 HL AHL 5.13
- Variable base, variable exponent. Functions like $y = x^{x}$ or $y = (\sin x)^{x}$. The power rule does not apply (the exponent is not constant), and the exponential-derivative rule does not apply (the base is not constant). Logarithmic differentiation handles both.
- Long products and quotients of powers. Functions like $y = \dfrac{(x + 1)^{4} (x - 2)^{3}}{(x^{2} + 1)^{5}}$. Multiple product and quotient rule applications get unwieldy; taking $\ln$ converts the product into a sum, which then differentiates one term at a time.
- Take $\ln$ of both sides: $\ln y = \ln \!\bigl[\text{expression}\bigr]$.
- Use log laws to expand the right side into a sum.
- Differentiate implicitly: the left side becomes $\dfrac{1}{y} \dfrac{dy}{dx}$.
- Multiply through by $y$ (substituting back the original expression).
- 底数与指数同时含 $x$。如 $y = x^{x}$、$y = (\sin x)^{x}$。幂法则不适用（指数非常数），指数法则也不适用（底数非常数）。对数求导可同时处理。
- 大量幂次相乘除。如 $y = \dfrac{(x + 1)^{4} (x - 2)^{3}}{(x^{2} + 1)^{5}}$。多次套用乘积与商法则会很笨重；取 $\ln$ 把乘积变为求和，再逐项求导。
- 两边取 $\ln$：$\ln y = \ln \!\bigl[\text{表达式}\bigr]$。
- 用对数律把右侧展开为求和。
- 隐式求导：左侧变为 $\dfrac{1}{y} \dfrac{dy}{dx}$。
- 两边同乘 $y$（再代回原表达式）。
Find $\dfrac{dy}{dx}$ where $y = x^{x}$ for $x > 0$.设 $y = x^{x}$（$x > 0$），求 $\dfrac{dy}{dx}$。
Take $\ln$.
取 $\ln$。
$$ \ln y \;=\; \ln(x^{x}) \;=\; x \ln x. $$Differentiate both sides with respect to $x$. Left side uses the chain rule; right side uses the product rule.
两边对 $x$ 求导。左侧用链式法则；右侧用乘积法则。
$$ \frac{1}{y} \frac{dy}{dx} \;=\; \ln x \,+\, x \cdot \frac{1}{x} \;=\; \ln x + 1. $$Solve for $\dfrac{dy}{dx}$ and substitute back $y = x^{x}$.
解出 $\dfrac{dy}{dx}$ 并代回 $y = x^{x}$。
$$ \frac{dy}{dx} \;=\; y \,(\ln x + 1) \;=\; x^{x} \,(\ln x + 1). $$Find $\dfrac{dy}{dx}$ where $y = \dfrac{(x + 1)^{4} (x - 2)^{3}}{(x^{2} + 1)^{5}}$.设 $y = \dfrac{(x + 1)^{4} (x - 2)^{3}}{(x^{2} + 1)^{5}}$，求 $\dfrac{dy}{dx}$。
Take $\ln$ and expand using log laws.
取 $\ln$，用对数律展开。
$$ \ln y \;=\; 4 \ln(x + 1) \,+\, 3 \ln(x - 2) \,-\, 5 \ln(x^{2} + 1). $$Differentiate term by term. Each $\ln$ contributes a $1/(\text{argument}) \cdot (\text{derivative of argument})$ factor.
逐项求导。每个 $\ln$ 贡献因子 $1/(\text{自变量}) \cdot (\text{自变量的导数})$。
$$ \frac{1}{y} \frac{dy}{dx} \;=\; \frac{4}{x + 1} \,+\, \frac{3}{x - 2} \,-\, \frac{10 x}{x^{2} + 1}. $$Multiply by $y$.
两边乘以 $y$。
$$ \frac{dy}{dx} \;=\; \frac{(x + 1)^{4} (x - 2)^{3}}{(x^{2} + 1)^{5}} \left[ \frac{4}{x + 1} \,+\, \frac{3}{x - 2} \,-\, \frac{10 x}{x^{2} + 1} \right]. $$Remark. The factor in brackets is the same shape every time: each linear factor of the numerator contributes a positive reciprocal, each factor of the denominator contributes a negative reciprocal, scaled by the original exponent. Once you see the pattern, you can write $\tfrac{dy}{dx}$ directly without re-deriving.
注。方括号内的形式有规律：分子每个线性因子贡献"正的倒数"，分母每个线性因子贡献"负的倒数"，缩放系数为原指数。看出规律后可直接写出 $\tfrac{dy}{dx}$ 而无需重新推导。
▸ Going deeper: why the trick works▸ 深入：该技巧为何成立
Logarithms turn products into sums and powers into products: $\ln(ab) = \ln a + \ln b$ and $\ln(a^{n}) = n \ln a$. Both operations make differentiation easier, because sums and constant multiples are the simplest things to differentiate. The cost is one extra implicit-differentiation step on the left side ($\tfrac{d}{dx} \ln y = \tfrac{1}{y} \tfrac{dy}{dx}$), but that is one operation in exchange for cleaning up the entire right side.
对数把乘积变求和、把幂变乘积：$\ln(ab) = \ln a + \ln b$、$\ln(a^{n}) = n \ln a$。两种运算都让求导更简单，因为求和与常数倍是最易求导的形式。代价仅在于左侧需多一次隐函数求导（$\tfrac{d}{dx} \ln y = \tfrac{1}{y} \tfrac{dy}{dx}$），但换来的是右侧整体的极大简化。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- Read the function structure before differentiating. A composite $f(g(x))$ needs the chain rule; a product $f(x) g(x)$ needs the product rule; a quotient $f(x)/g(x)$ needs the quotient rule. Choose first, compute second.
- 动笔前先看函数结构。复合形 $f(g(x))$ 用链式法则；乘积形 $f(x) g(x)$ 用乘积法则；商形 $f(x)/g(x)$ 用商法则。先选规则，再计算。
- Combine rules when needed. Many problems are products of composites or quotients with chains inside. Layer the rules: outer rule first, then inner derivatives, each computed with whichever rule fits.
- 多条规则可叠加使用。许多题目是"复合的乘积"或"含链式的商"。从外到内依次套用各条规则。
- Simplify before solving for tangent slope. If you need $f'(a)$ at a single point, plug in $x = a$ before doing algebraic simplification of $f'(x)$. Often the substitution makes most terms vanish.
- 求点处切线斜率前先简化。若只需 $f'(a)$ 单点值，可在化简 $f'(x)$ 之前直接代 $x = a$。多数项往往因代入而归零。
- Mark every $y$ with a chain factor. The single most common deduction is omitting $\tfrac{dy}{dx}$ on a $y$-term. Before collecting, scan: every $y$-containing term should carry $\tfrac{dy}{dx}$ exactly once.
- 每个 $y$ 都要附链式因子。最常见扣分项是漏写 $\tfrac{dy}{dx}$。整理前先扫一遍：每个含 $y$ 的项都应恰好带一个 $\tfrac{dy}{dx}$。
- The product rule on $xy$ produces two pieces. $\tfrac{d}{dx}(xy) = y + x \tfrac{dy}{dx}$. Forgetting either piece costs A1.
- 对 $xy$ 用乘积法则会产生两项。$\tfrac{d}{dx}(xy) = y + x \tfrac{dy}{dx}$。漏写任一项会扣 A1。
- Always solve for $\tfrac{dy}{dx}$ explicitly. An implicit derivative left in implicit form does not earn the final A1. Collect, factor, and divide.
- 必须显式解出 $\tfrac{dy}{dx}$。停留在隐式形式拿不到最后 A1。整理、提因、除式都要做完。
- When the base depends on $x$ and so does the exponent, this is your only option. Power rule and exponential rule both fail. Reach for $\ln$.
- 当底数与指数都依赖 $x$ 时，对数求导是唯一选择。幂法则与指数法则都失效。直接取 $\ln$。
- Expand fully before differentiating. Use $\ln(ab) = \ln a + \ln b$ and $\ln(a^{n}) = n \ln a$ to flatten the right side into a sum. Differentiating a sum is much easier than differentiating a nested expression.
- 求导前先充分展开。用 $\ln(ab) = \ln a + \ln b$、$\ln(a^{n}) = n \ln a$ 把右侧展平为求和。求和的导数比嵌套表达式简单得多。
- Remember to multiply back by $y$. The intermediate result is $\tfrac{1}{y} \tfrac{dy}{dx}$, not $\tfrac{dy}{dx}$ itself. Multiplying by $y$ and substituting back the original expression is the final A1.
- 记得最后乘回 $y$。中间结果是 $\tfrac{1}{y} \tfrac{dy}{dx}$，而非 $\tfrac{dy}{dx}$ 本身。乘以 $y$ 并代回原表达式才是最后的 A1。
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Unit E2 Practice Quiz单元 E2 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, without the formula box, on your first attempt.
每一条都要"裸做"做对（不看笔记、不看公式框、一次过）才打勾。
- Apply the power rule to polynomials with negative and fractional exponents对含负指数与分数指数的多项式使用幂法则
- State the chain rule formula from memory and apply it to a single-nested composite凭记忆写出链式法则并应用于单层嵌套复合函数
- Apply the chain rule to a double-nested composite such as $\sin(\ln(x^{2} + 1))$将链式法则应用于双层嵌套（如 $\sin(\ln(x^{2} + 1))$）
- Apply the product rule correctly in both orderings在两种次序下都能正确使用乘积法则
- Apply the quotient rule with the correct minus-sign ordering ($u' v - u v'$ in the numerator)使用商法则时分子保持正确的减号次序（$u' v - u v'$）
- Reproduce the standard derivatives of $\sin$, $\cos$, $\tan$, $e^{x}$, $\ln x$, $a^{x}$复现 $\sin$、$\cos$、$\tan$、$e^{x}$、$\ln x$、$a^{x}$ 的标准导数
- Combine the chain rule with the standard trig, exp, and log derivatives把链式法则与标准三角、指数、对数导数结合
- HL Reproduce the inverse trig derivatives ($\arcsin$, $\arccos$, $\arctan$)复现反三角函数（$\arcsin$、$\arccos$、$\arctan$）的导数
- HL Differentiate implicitly, including curves with mixed $xy$ terms requiring the product rule做隐函数求导，含 $xy$ 混合项（需乘积法则）的情形
- HL Solve for $\tfrac{dy}{dx}$ explicitly after implicit differentiation隐式求导后显式解出 $\tfrac{dy}{dx}$
- HL Apply logarithmic differentiation to a variable-base, variable-exponent function (e.g. $x^{x}$)对底数与指数都含 $x$ 的函数（如 $x^{x}$）使用对数求导
- HL Apply logarithmic differentiation to a long product or quotient of powers对幂次相乘除的长式使用对数求导
- Recognise which of the six rules applies by inspecting the structure of the function通过观察函数结构辨别六条规则中应当使用哪一条
IB Paper-Style PracticeIB 试卷风格练习
E2 Practice and Solutions are on the roadmap (sibling of the A and D Practice sets). They will ship under Practice Questions/Unit_E2_*.html with the bilingual built-in pattern. In the meantime, the A2 and A5 Practice sets demonstrate the format.
E2 配套的 Practice 与 Solutions 已在排期（与 A 系列、D 系列的练习集同级）。上线后将位于 Practice Questions/Unit_E2_*.html，采用双语内嵌格式。在此之前，A2 与 A5 的练习集展示了相同格式。