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Unit E2 · SolutionsUnit E2 · 解析

Techniques of Differential Calculus · Solutions微分技巧 · 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus SL 5.3, 5.6, 5.7 · AHL 5.13考纲 SL 5.3、5.6、5.7 · AHL 5.13AA HL

PART I  ·  PAPER 1 SECTION A · SOLUTIONS第一部分  ·  第一卷 A 节 · 解析No calculator · 21 marks不可使用计算器 · 21 分

Section A · Worked SolutionsA 节 · 详细解析

Q1EASYPaper 1ASL 5.7 Chain Rule · Square Root[4 marks]

$f(x) = \sqrt{1 + x^{4}}$ for $x \in \mathbb{R}$. (a) Write as $u^{1/2}$, identify $u$. (b) Apply chain rule to find $f'(x)$ and simplify.$f(x) = \sqrt{1 + x^{4}}$，$x \in \mathbb{R}$。(a) 写成 $u^{1/2}$ 形式并指出 $u$。(b) 用链式法则求 $f'(x)$ 并化简。

Answer:答案：  $f'(x) = \dfrac{2 x^{3}}{\sqrt{1 + x^{4}}}$

(a) Recognise the composition A1

Set $u = 1 + x^{4}$. Then $f(x) = u^{1/2}$, the outer function being $u \mapsto u^{1/2}$ and the inner function $u(x) = 1 + x^{4}$.

(b) Chain rule M1·A1·A1

$\dfrac{du}{dx} = 4 x^{3}$. The chain rule gives $$ f'(x) \;=\; \tfrac{1}{2}\, u^{-1/2} \cdot \tfrac{du}{dx} \;=\; \tfrac{1}{2}\,(1 + x^{4})^{-1/2} \cdot 4 x^{3} \;=\; \frac{2 x^{3}}{\sqrt{1 + x^{4}}}. $$
The $\tfrac{1}{2} \cdot 4 = 2$ collapse is where marks live. Many students stop at $\tfrac{1}{2}(1 + x^{4})^{-1/2} \cdot 4 x^{3}$ and forget to simplify. IB markschemes reserve the final A1 for the simplified form. Two safe simplification rules: (i) negative fractional exponents go to the denominator as a radical: $u^{-1/2} = 1/\sqrt{u}$; (ii) numerical coefficients are always combined: $\tfrac{1}{2} \cdot 4 = 2$. Train the eye to write the simplified line on the next row, never inside the same expression.

(a) 识别复合结构 A1

令 $u = 1 + x^{4}$。则 $f(x) = u^{1/2}$，外函数为 $u \mapsto u^{1/2}$，内函数为 $u(x) = 1 + x^{4}$。

(b) 链式法则 M1·A1·A1

$\dfrac{du}{dx} = 4 x^{3}$。由链式法则： $$ f'(x) \;=\; \tfrac{1}{2}\, u^{-1/2} \cdot \tfrac{du}{dx} \;=\; \tfrac{1}{2}\,(1 + x^{4})^{-1/2} \cdot 4 x^{3} \;=\; \frac{2 x^{3}}{\sqrt{1 + x^{4}}}. $$
$\tfrac{1}{2} \cdot 4 = 2$ 这一步合并就是丢分点。很多学生停在 $\tfrac{1}{2}(1 + x^{4})^{-1/2} \cdot 4 x^{3}$ 不再化简。IB 评分标准把最后一个 A1 留给化简形式。两条稳妥的化简规则：(i) 负分数指数移到分母变根号：$u^{-1/2} = 1/\sqrt{u}$；(ii) 数字系数务必合并：$\tfrac{1}{2} \cdot 4 = 2$。训练自己另起一行写化简结果，不要塞在同一行表达式内。
Q2MEDIUMPaper 1ASL 5.7 Product Rule · Trig Pairing[5 marks]

$y = x^{2} \sin x$. (a) State the product rule. (b) Find $\dfrac{dy}{dx}$ and simplify. (c) Evaluate at $x = \pi$.$y = x^{2} \sin x$。(a) 写出乘积法则。(b) 求 $\dfrac{dy}{dx}$ 并化简。(c) 在 $x = \pi$ 处求值。

Answer:答案：  (b) $\dfrac{dy}{dx} = 2 x \sin x + x^{2} \cos x$  ·  (c) $-\pi^{2}$

(a) Product rule A1

For $y = u(x)\,v(x)$: $\;\dfrac{dy}{dx} = u'(x)\,v(x) + u(x)\,v'(x)$.

(b) Apply with $u = x^{2},\; v = \sin x$ M1·A1·A1

$u' = 2 x$ and $v' = \cos x$. Thus $$ \frac{dy}{dx} \;=\; 2 x \cdot \sin x + x^{2} \cdot \cos x \;=\; 2 x \sin x + x^{2} \cos x. $$ Factor optional: $x\,(2 \sin x + x \cos x)$.

(c) Evaluate at $x = \pi$ A1

$\sin \pi = 0$, $\cos \pi = -1$. So $\;\dfrac{dy}{dx}\bigg|_{x=\pi} = 2\pi \cdot 0 + \pi^{2}\cdot(-1) = -\pi^{2}$.
The order matters for partial credit. Markschemes credit "$u' v$" and "$u v'$" as separate A1s. If you collapse them straight into $2 x \sin x + x^{2} \cos x$ with no intermediate line, an A1 is at risk. Standard exam habit: write the rule template first (the A1 in part (a)), then substitute the four pieces $(u, u', v, v')$ on one line, then the simplified sum on the next. The structure protects each mark and signals to the marker that you applied the rule, not memorised the answer.

(a) 乘积法则 A1

对 $y = u(x)\,v(x)$：$\;\dfrac{dy}{dx} = u'(x)\,v(x) + u(x)\,v'(x)$。

(b) 取 $u = x^{2},\; v = \sin x$ M1·A1·A1

$u' = 2 x$，$v' = \cos x$。故 $$ \frac{dy}{dx} \;=\; 2 x \cdot \sin x + x^{2} \cdot \cos x \;=\; 2 x \sin x + x^{2} \cos x. $$ 可选提公因式：$x\,(2 \sin x + x \cos x)$。

(c) $x = \pi$ 处求值 A1

$\sin \pi = 0$，$\cos \pi = -1$。故 $\;\dfrac{dy}{dx}\bigg|_{x=\pi} = 2\pi \cdot 0 + \pi^{2}\cdot(-1) = -\pi^{2}$。
顺序影响过程分。评分标准把 "$u' v$" 与 "$u v'$" 视为各自的 A1。如果直接合写成 $2 x \sin x + x^{2} \cos x$ 不留中间步骤，会丢一个 A1。规范考试动作：先写法则模板（(a) 部分的 A1），再一行代入四个量 $(u, u', v, v')$，再下一行写出化简和。这个结构保住每一分，也让评卷人看到你"用了法则"，而非"背了答案"。
Q3MEDIUMPaper 1ASL 5.7 Quotient Rule · Exponential[6 marks]

$y = \dfrac{e^{x}}{x}$, $x \ne 0$. (a) Quotient rule. (b) Find $\dfrac{dy}{dx}$, factor $e^{x}$. (c) Stationary points and classify.$y = \dfrac{e^{x}}{x}$，$x \ne 0$。(a) 商法则。(b) 求 $\dfrac{dy}{dx}$ 并提出 $e^{x}$。(c) 驻点并分类。

Answer:答案：  (b) $\dfrac{dy}{dx} = \dfrac{e^{x}(x - 1)}{x^{2}}$  ·  (c) $(1,\, e)$ local minimum

(a) Quotient rule A1

For $y = \dfrac{u(x)}{v(x)}$: $\;\dfrac{dy}{dx} = \dfrac{u'(x)\,v(x) - u(x)\,v'(x)}{[v(x)]^{2}}$.

(b) Apply with $u = e^{x},\; v = x$ M1·A1·A1

$u' = e^{x}$, $v' = 1$. Hence $$ \frac{dy}{dx} \;=\; \frac{e^{x} \cdot x - e^{x} \cdot 1}{x^{2}} \;=\; \frac{e^{x}(x - 1)}{x^{2}}. $$

(c) Stationary points M1·A1

Set $\dfrac{dy}{dx} = 0$. Since $e^{x} > 0$ and $x^{2} > 0$ (on the domain $x \ne 0$), the only zero is $x - 1 = 0$, i.e. $x = 1$. Value: $y(1) = e^{1}/1 = e$. Classification by sign of $x - 1$ on either side:
  • $x \in (0, 1)$: $x - 1 < 0 \Rightarrow y' < 0$ (decreasing).
  • $x \in (1, \infty)$: $x - 1 > 0 \Rightarrow y' > 0$ (increasing).
So $x = 1$ is a local minimum at $(1, e)$.
Numerator-only sign analysis on $\mathbb{R}^{+}$. A frequent IB shortcut: when a derivative is $e^{x} \cdot (\text{polynomial}) / x^{n}$ with $x > 0$, the sign of $\dfrac{dy}{dx}$ equals the sign of the polynomial. You do not need the full second-derivative test. Write the line "since $e^{x} > 0$ and $x^{2} > 0$, $\operatorname{sign}(y') = \operatorname{sign}(x - 1)$" and you have earned the R1 for justification while skipping algebra. Note also: the branch $x < 0$ has no stationary point but is monotonic (the same sign argument: $y' = e^{x}(x-1)/x^{2}$ is negative there since $x - 1 < 0$), so the function decreases on $(-\infty, 0)$.

(a) 商法则 A1

对 $y = \dfrac{u(x)}{v(x)}$：$\;\dfrac{dy}{dx} = \dfrac{u'(x)\,v(x) - u(x)\,v'(x)}{[v(x)]^{2}}$。

(b) 取 $u = e^{x},\; v = x$ M1·A1·A1

$u' = e^{x}$，$v' = 1$。故 $$ \frac{dy}{dx} \;=\; \frac{e^{x} \cdot x - e^{x} \cdot 1}{x^{2}} \;=\; \frac{e^{x}(x - 1)}{x^{2}}. $$

(c) 驻点 M1·A1

令 $\dfrac{dy}{dx} = 0$。由 $e^{x} > 0$ 与 $x^{2} > 0$（$x \ne 0$），仅 $x - 1 = 0$，即 $x = 1$。代入 $y(1) = e^{1}/1 = e$。两侧符号分析：
  • $x \in (0, 1)$：$x - 1 < 0 \Rightarrow y' < 0$（递减）。
  • $x \in (1, \infty)$：$x - 1 > 0 \Rightarrow y' > 0$（递增）。
故 $x = 1$ 为极小值点，坐标 $(1, e)$。
$\mathbb{R}^{+}$ 上只看分子的符号。IB 常用捷径：当导数形如 $e^{x} \cdot p(x) / x^{n}$（$x > 0$，$p$ 为多项式）时，$\dfrac{dy}{dx}$ 的符号即 $p(x)$ 的符号。无须二阶导分类。写一行"因 $e^{x} > 0$、$x^{2} > 0$，故 $\operatorname{sign}(y') = \operatorname{sign}(x - 1)$"，即获 R1 论证分且省去代数。另外：$x < 0$ 分支无驻点但单调（同样符号论证：$y' = e^{x}(x-1)/x^{2}$ 在那里为负，因 $x - 1 < 0$），故函数在 $(-\infty, 0)$ 上递减。
Q4HARDPaper 1ASL 5.7 Chain Rule · Log Composition[6 marks]

$g(x) = \ln(\cos x)$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. (a) Well-defined? (b) Show $g'(x) = -\tan x$. (c) Find $g''(x)$, evaluate at $0$.$g(x) = \ln(\cos x)$，定义于 $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$。(a) 是否有定义？(b) 证明 $g'(x) = -\tan x$。(c) 求 $g''(x)$，并在 $x = 0$ 处求值。

Answer:答案：  (b) $g'(x) = -\tan x$ (AG)  ·  (c) $g''(x) = -\sec^{2} x$, $g''(0) = -1$

(a) Well-defined R1

For $\ln(\cdot)$ to be defined we need $\cos x > 0$. On $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, $\cos x > 0$ throughout (it equals $1$ at $0$, decreases to $0$ at the endpoints but never reaches $0$ inside). So $g$ is well-defined on this open interval.

(b) Chain rule on $\ln(\cos x)$ M1·A1·A1

Outer $u \mapsto \ln u$ with derivative $1/u$; inner $u = \cos x$ with derivative $-\sin x$. Hence $$ g'(x) \;=\; \frac{1}{\cos x} \cdot (-\sin x) \;=\; -\frac{\sin x}{\cos x} \;=\; -\tan x. \quad \text{AG} $$

(c) Differentiate $-\tan x$ M1·A1

$\dfrac{d}{dx}\tan x = \sec^{2} x$ (standard derivative), so $g''(x) = -\sec^{2} x$. At $x = 0$: $\sec 0 = 1/\cos 0 = 1$, hence $g''(0) = -1$.
Sign of $g''$ as a concavity tell. $g''(x) = -\sec^{2} x < 0$ for all $x$ in the open interval, so $g(x) = \ln(\cos x)$ is strictly concave on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. This makes the function its own "second-order test cheat sheet": no need to compute $g'(0) = -\tan 0 = 0$ and $g''(0) = -1 < 0$ separately — together they certify a local maximum at $x = 0$, which is the unique critical point. Visualisation: $\ln(\cos x)$ is the bell whose graph passes through $(0, 0)$ and falls off to $-\infty$ at the asymptotes $x = \pm \pi/2$. Concavity throughout the interval is a frequent IB hidden-marks bonus when the question asks "comment on the shape".

(a) 有定义性 R1

$\ln(\cdot)$ 有定义需 $\cos x > 0$。在 $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ 内 $\cos x > 0$（在 $0$ 处为 $1$，向端点递减至 $0$，但内部不取 $0$）。故 $g$ 在该开区间上有定义。

(b) 对 $\ln(\cos x)$ 用链式法则 M1·A1·A1

外函数 $u \mapsto \ln u$，导数 $1/u$；内函数 $u = \cos x$，导数 $-\sin x$。故 $$ g'(x) \;=\; \frac{1}{\cos x} \cdot (-\sin x) \;=\; -\frac{\sin x}{\cos x} \;=\; -\tan x. \quad \text{AG} $$

(c) 对 $-\tan x$ 求导 M1·A1

$\dfrac{d}{dx}\tan x = \sec^{2} x$（标准结果），故 $g''(x) = -\sec^{2} x$。在 $x = 0$ 处：$\sec 0 = 1/\cos 0 = 1$，故 $g''(0) = -1$。
$g''$ 的符号即凹性提示。$g''(x) = -\sec^{2} x < 0$ 在开区间上恒成立，故 $g(x) = \ln(\cos x)$ 在 $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ 上严格凹。这相当于函数自带"二阶分类备忘"：无须分别算 $g'(0) = -\tan 0 = 0$ 与 $g''(0) = -1 < 0$ —— 两者合起来即可断定 $x = 0$ 为唯一极大值点。形象：$\ln(\cos x)$ 是过 $(0, 0)$ 的"钟形"，两端 $x = \pm \pi/2$ 处渐近至 $-\infty$。当题目问"说明图像形状"时，全区间凹性常是 IB 的隐藏加分点。
PART II  ·  PAPER 1 SECTION B · SOLUTIONS第二部分  ·  第一卷 B 节 · 解析No calculator · 11 marks不可使用计算器 · 11 分

Section B · Worked SolutionsB 节 · 详细解析

Q5HARDPaper 1BAHL 5.13 Implicit Differentiation (HL)[11 marks]

Curve $C$: $x^{3} + 3xy + y^{3} = 1$. (a) $P(0,1)$ on $C$. (b) Differentiate. (c) Show $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$, state undefined points. (d) Tangent at $P$ as $y = mx + c$. (e) As $ax + by = c$ with integer coefficients.曲线 $C$：$x^{3} + 3xy + y^{3} = 1$。(a) $P(0,1)$ 在 $C$ 上。(b) 求导。(c) 证 $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$，指出无意义点。(d) $P$ 处切线 $y = mx + c$。(e) 整系数 $ax + by = c$。

Answers:答案：  (c) undefined iff $x + y^{2} = 0$  ·  (d) $y = -x + 1$  ·  (e) $x + y = 1$

(a) Verify $P$ on $C$ A1

$0^{3} + 3(0)(1) + 1^{3} = 0 + 0 + 1 = 1$. $\checkmark$

(b) Implicit differentiation M1·A1·A1

Differentiate $x^{3} + 3xy + y^{3} = 1$ with respect to $x$, treating $y$ as a function of $x$: $$ 3 x^{2} + 3\bigl(y + x\,\tfrac{dy}{dx}\bigr) + 3 y^{2}\,\tfrac{dy}{dx} \;=\; 0. $$ The middle term uses the product rule on $3xy$; the last uses the chain rule on $y^{3}$.

(c) Solve for $\dfrac{dy}{dx}$ M1·A1·A1

Divide both sides by $3$ and group: $$ x^{2} + y + (x + y^{2})\,\tfrac{dy}{dx} \;=\; 0 \;\Longrightarrow\; (x + y^{2})\,\tfrac{dy}{dx} \;=\; -(x^{2} + y). $$ Hence $\;\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$. Undefined when $x + y^{2} = 0$, i.e. on the locus $x = -y^{2}$ (intersected with $C$ — these are the points with vertical tangents).

(d) Tangent at $P(0, 1)$ M1·A1

$\;\dfrac{dy}{dx}\bigg|_{(0,1)} = -\dfrac{0 + 1}{0 + 1} = -1$. Point-slope: $y - 1 = -1(x - 0)$, i.e. $y = -x + 1$.

(e) Integer-coefficient form A1·A1

Rearrange $y = -x + 1$ to $x + y = 1$. Coefficients $(1, 1, 1)$ are integers.
Vertical-tangent locus is hidden geometry. The expression $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$ is undefined exactly when the implicit curve has a vertical tangent (slope $= \infty$, i.e. the "$y$ on $x$" view fails). Setting $x + y^{2} = 0$ gives $x = -y^{2}$; substituting into the curve equation: $(-y^{2})^{3} + 3(-y^{2})y + y^{3} = -y^{6} - 3 y^{3} + y^{3} = -y^{6} - 2 y^{3} = 1$, so $y^{6} + 2 y^{3} + 1 = 0$, i.e. $(y^{3} + 1)^{2} = 0$, giving $y = -1$ and $x = -1$. The single point $(-1, -1)$ on $C$ has a vertical tangent. Knowing this kind of hidden structure turns a routine HL implicit problem into a Paper 3 exploration target.

(a) 验证 $P$ 在 $C$ 上 A1

$0^{3} + 3(0)(1) + 1^{3} = 0 + 0 + 1 = 1$。$\checkmark$

(b) 隐函数求导 M1·A1·A1

把 $y$ 视为 $x$ 的函数，对 $x^{3} + 3xy + y^{3} = 1$ 两边求导： $$ 3 x^{2} + 3\bigl(y + x\,\tfrac{dy}{dx}\bigr) + 3 y^{2}\,\tfrac{dy}{dx} \;=\; 0. $$ 中间项用乘积法则求 $3xy$，末项用链式法则求 $y^{3}$。

(c) 解出 $\dfrac{dy}{dx}$ M1·A1·A1

两边除 $3$ 并归项： $$ x^{2} + y + (x + y^{2})\,\tfrac{dy}{dx} \;=\; 0 \;\Longrightarrow\; (x + y^{2})\,\tfrac{dy}{dx} \;=\; -(x^{2} + y). $$ 故 $\;\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$。当 $x + y^{2} = 0$（即 $x = -y^{2}$，并与 $C$ 联立的点）时无意义——即出现竖直切线的位置。

(d) $P(0, 1)$ 处切线 M1·A1

$\;\dfrac{dy}{dx}\bigg|_{(0,1)} = -\dfrac{0 + 1}{0 + 1} = -1$。点斜式：$y - 1 = -1(x - 0)$，即 $y = -x + 1$。

(e) 整系数形式 A1·A1

把 $y = -x + 1$ 整理为 $x + y = 1$。系数 $(1, 1, 1)$ 皆为整数。
竖直切线的轨迹是隐藏几何。表达式 $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$ 恰在曲线出现竖直切线（斜率为 $\infty$，"$y$ 关于 $x$"视角失效）时无意义。令 $x + y^{2} = 0$ 得 $x = -y^{2}$；代回曲线方程：$(-y^{2})^{3} + 3(-y^{2})y + y^{3} = -y^{6} - 2 y^{3} = 1$，即 $(y^{3} + 1)^{2} = 0$，得 $y = -1$、$x = -1$。曲线上唯一具有竖直切线的点为 $(-1, -1)$。掌握这种隐藏结构，能把 HL 例行隐函数题升级为 Paper 3 探究目标。
PART III  ·  PAPER 2 · SOLUTIONS第三部分  ·  第二卷 · 解析Calculator · 16 marks可使用计算器 · 16 分

Paper 2 · Worked Solutions第二卷 · 详细解析

Q6MEDIUMPaper 2SL 5.7 Tangent Slopes · GDC Confirmation[7 marks]

$f(x) = x\, e^{-x^{2}/2}$ on $\mathbb{R}$. (a) Show $f'(x) = (1 - x^{2})\,e^{-x^{2}/2}$. (b) Stationary points exactly + GDC. (c) $f'(0.5)$ to 3 s.f., verify.$f(x) = x\, e^{-x^{2}/2}$（$\mathbb{R}$）。(a) 证明 $f'(x) = (1 - x^{2})\,e^{-x^{2}/2}$。(b) 精确驻点 + GDC。(c) $f'(0.5)$（3 sf），核对。

Answers:答案：  (b) $x = \pm 1$  ·  (c) $f'(0.5) \approx 0.662$

(a) Product + chain rule M1·A1·A1

With $u = x$, $v = e^{-x^{2}/2}$: $u' = 1$ and $v' = e^{-x^{2}/2} \cdot \dfrac{d}{dx}\bigl(-\tfrac{x^{2}}{2}\bigr) = -x\,e^{-x^{2}/2}$ by the chain rule. Therefore $$ f'(x) \;=\; 1 \cdot e^{-x^{2}/2} + x \cdot (-x\,e^{-x^{2}/2}) \;=\; e^{-x^{2}/2}(1 - x^{2}). \quad \text{AG} $$

(b) Stationary points M1·A1

$f'(x) = 0 \Longleftrightarrow 1 - x^{2} = 0$ (since $e^{-x^{2}/2} > 0$ always), so $x = \pm 1$. GDC confirmation: graph $y_{1} = f(x)$ and $y_{2} = f'(x)$; observe $f'$ crosses zero at $x = \pm 1$ and $f$ has local extrema there (max at $x = 1$ with $f(1) = e^{-1/2} \approx 0.607$, min at $x = -1$ with $f(-1) \approx -0.607$).

(c) Numerical check at $x = 0.5$ M1·A1

GDC nDeriv: $f'(0.5) \approx 0.6620$. Closed form: $f'(0.5) = (1 - 0.25)\,e^{-0.125} = 0.75 \cdot e^{-0.125} \approx 0.75 \cdot 0.8825 \approx 0.6619$, agreeing to 3 s.f. as $0.662$.
Why $f(x) = x\,e^{-x^{2}/2}$ is a stealth normal-distribution question. The function is (up to a constant) the derivative of the standard normal density $\phi(x) = \tfrac{1}{\sqrt{2\pi}}\, e^{-x^{2}/2}$: $\phi'(x) = -x\, \phi(x) = -x\, e^{-x^{2}/2} / \sqrt{2\pi}$. So $f(x)$ is $-\sqrt{2\pi}\,\phi'(x)$. The stationary points of $f$ at $x = \pm 1$ correspond to the inflection points of $\phi$ — exactly where the normal curve changes concavity, which is one standard deviation from the mean. IB Paper 3 explorations frequently fuse calculus and statistics through this exact identity: knowing it makes the "why $\pm 1$?" question into a structural recognition rather than algebra.

(a) 乘积 + 链式法则 M1·A1·A1

取 $u = x$、$v = e^{-x^{2}/2}$：$u' = 1$；由链式法则 $v' = e^{-x^{2}/2} \cdot \dfrac{d}{dx}\bigl(-\tfrac{x^{2}}{2}\bigr) = -x\,e^{-x^{2}/2}$。故 $$ f'(x) \;=\; 1 \cdot e^{-x^{2}/2} + x \cdot (-x\,e^{-x^{2}/2}) \;=\; e^{-x^{2}/2}(1 - x^{2}). \quad \text{AG} $$

(b) 驻点 M1·A1

$f'(x) = 0 \Longleftrightarrow 1 - x^{2} = 0$（因 $e^{-x^{2}/2} > 0$ 恒成立），故 $x = \pm 1$。GDC 核对：绘 $y_{1} = f(x)$ 与 $y_{2} = f'(x)$；$f'$ 在 $x = \pm 1$ 处过零，$f$ 在该处取极值（$x = 1$ 极大 $f(1) = e^{-1/2} \approx 0.607$，$x = -1$ 极小 $\approx -0.607$）。

(c) $x = 0.5$ 处数值核对 M1·A1

GDC nDeriv：$f'(0.5) \approx 0.6620$。闭式：$f'(0.5) = (1 - 0.25)\,e^{-0.125} = 0.75 \cdot e^{-0.125} \approx 0.75 \cdot 0.8825 \approx 0.6619$，与 GDC 一致，$3$ 位有效数字为 $0.662$。
$f(x) = x\,e^{-x^{2}/2}$ 是隐形的正态分布题。该函数（差一常数）正是标准正态密度 $\phi(x) = \tfrac{1}{\sqrt{2\pi}}\, e^{-x^{2}/2}$ 的导数：$\phi'(x) = -x\, \phi(x) = -x\, e^{-x^{2}/2} / \sqrt{2\pi}$。故 $f(x) = -\sqrt{2\pi}\,\phi'(x)$。$f$ 在 $x = \pm 1$ 的驻点对应 $\phi$ 的拐点——正态曲线凹性改变的位置，恰为离均值一个标准差处。Paper 3 探究常把微积分与统计藉此恒等式融合：认出这一身份后，"为什么是 $\pm 1$？"就从代数变为结构识别。
Q7HARDPaper 2AHL 5.13 Tangent to Implicit Curve (HL)[9 marks]

$K$: $x^{2} + xy + y^{2} = 3$. (a) $Q(1,1)$ on $K$. (b) Implicit differentiation. (c) $\left.\dfrac{dy}{dx}\right|_{(1,1)}$ + tangent at $Q$. (d) Confirm with GDC.$K$：$x^{2} + xy + y^{2} = 3$。(a) $Q(1,1)$ 在 $K$ 上。(b) 隐函数求导。(c) $\left.\dfrac{dy}{dx}\right|_{(1,1)}$ + $Q$ 处切线。(d) GDC 验证。

Answers:答案：  (c) slope $= -1$, tangent $y = -x + 2$

(a) Verify $Q$ on $K$ A1

$1^{2} + (1)(1) + 1^{2} = 1 + 1 + 1 = 3$. $\checkmark$

(b) Implicit differentiation M1·A1·A1

Differentiate $x^{2} + xy + y^{2} = 3$ w.r.t. $x$, $y = y(x)$: $$ 2 x + \bigl(y + x\,\tfrac{dy}{dx}\bigr) + 2 y\,\tfrac{dy}{dx} \;=\; 0. $$ The $xy$-term used the product rule; the $y^{2}$-term used the chain rule.

(c) Solve and substitute M1·A1·A1

Group $\dfrac{dy}{dx}$ terms: $$ (x + 2 y)\,\tfrac{dy}{dx} \;=\; -(2 x + y) \;\Longrightarrow\; \tfrac{dy}{dx} \;=\; -\dfrac{2 x + y}{x + 2 y}. $$ At $(1, 1)$: $\;\dfrac{dy}{dx} = -\dfrac{2 + 1}{1 + 2} = -\dfrac{3}{3} = -1$. Tangent at $Q$ through $(1, 1)$ with slope $-1$: $y - 1 = -1(x - 1)$, i.e. $y = -x + 2$ (or $x + y = 2$).

(d) GDC confirmation M1·A1

Plot $K$ implicitly (or as the two branches $y = \tfrac{1}{2}(-x \pm \sqrt{12 - 3 x^{2}})$). The upper branch passes through $(1, 1)$. Overlay the line $y = -x + 2$: it touches $K$ at $(1, 1)$ with the curve to one side, confirming slope $-1$.
The ellipse $x^{2} + xy + y^{2} = 3$ has a $45^{\circ}$-rotated axis. Although the equation looks asymmetric due to the $xy$ cross-term, the curve is a rotated ellipse: substitute $u = (x + y)/\sqrt{2}$, $v = (x - y)/\sqrt{2}$ (a $45^{\circ}$ rotation). Then $x^{2} + y^{2} = u^{2} + v^{2}$ and $xy = (u^{2} - v^{2})/2$, so the equation becomes $\tfrac{3}{2} u^{2} + \tfrac{1}{2} v^{2} = 3$, i.e. $u^{2}/2 + v^{2}/6 = 1$. This is an ellipse with semi-axes $\sqrt{2}$ and $\sqrt{6}$, rotated $45^{\circ}$. The tangent at $Q(1, 1)$, which is the point $(u, v) = (\sqrt{2}, 0)$ — the major-axis vertex in the $u v$-frame — must be vertical in those rotated coordinates, i.e. perpendicular to the $u$-axis. Perpendicular to $(1, 1)/\sqrt{2}$ is $(1, -1)/\sqrt{2}$, which has slope $-1$. The implicit derivative just rediscovered the rotated geometry.

(a) 验证 $Q$ 在 $K$ 上 A1

$1^{2} + (1)(1) + 1^{2} = 1 + 1 + 1 = 3$。$\checkmark$

(b) 隐函数求导 M1·A1·A1

把 $y = y(x)$，对 $x^{2} + xy + y^{2} = 3$ 两边求导： $$ 2 x + \bigl(y + x\,\tfrac{dy}{dx}\bigr) + 2 y\,\tfrac{dy}{dx} \;=\; 0. $$ $xy$ 项用乘积法则，$y^{2}$ 项用链式法则。

(c) 解出并代入 M1·A1·A1

归并 $\dfrac{dy}{dx}$ 项： $$ (x + 2 y)\,\tfrac{dy}{dx} \;=\; -(2 x + y) \;\Longrightarrow\; \tfrac{dy}{dx} \;=\; -\dfrac{2 x + y}{x + 2 y}. $$ 在 $(1, 1)$ 处：$\;\dfrac{dy}{dx} = -\dfrac{2 + 1}{1 + 2} = -1$。 $Q$ 处切线：$y - 1 = -1(x - 1)$，即 $y = -x + 2$（亦即 $x + y = 2$）。

(d) GDC 验证 M1·A1

用 GDC 隐式绘 $K$（或两支 $y = \tfrac{1}{2}(-x \pm \sqrt{12 - 3 x^{2}})$）。上支过 $(1, 1)$。叠加直线 $y = -x + 2$：在 $(1, 1)$ 处与 $K$ 相切，曲线全在一侧，斜率 $-1$ 得证。
椭圆 $x^{2} + xy + y^{2} = 3$ 有一条 $45^{\circ}$ 旋转的主轴。方程因含 $xy$ 交叉项看似不对称，但实为旋转椭圆：代换 $u = (x + y)/\sqrt{2}$、$v = (x - y)/\sqrt{2}$（$45^{\circ}$ 旋转）。则 $x^{2} + y^{2} = u^{2} + v^{2}$、$xy = (u^{2} - v^{2})/2$，方程化为 $\tfrac{3}{2} u^{2} + \tfrac{1}{2} v^{2} = 3$，即 $u^{2}/2 + v^{2}/6 = 1$。半长轴 $\sqrt{6}$、半短轴 $\sqrt{2}$，倾角 $45^{\circ}$。$Q(1, 1)$ 在 $uv$ 系下为 $(u, v) = (\sqrt{2}, 0)$——短轴顶点，切线在旋转坐标中垂直于 $u$ 轴。垂直于 $(1, 1)/\sqrt{2}$ 的方向为 $(1, -1)/\sqrt{2}$，斜率 $-1$。隐函数导数正是把旋转几何重新发现一遍。
PART IV  ·  PAPER 3 · SOLUTIONS第四部分  ·  第三卷 · 解析Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分

Paper 3 · Worked Solutions第三卷 · 详细解析

Q8HARDPaper 3AHL 5.13 Logarithmic Differentiation (HL)[15 marks]

Logarithmic differentiation of $y = u(x)^{v(x)}$. (a) $y = x^{\sin x}$ derivative. (b) Evaluate at $\pi/2$. (c) $z = (x^{2} + 1)^{x^{2}}$. (d) General formula and consistency.对 $y = u(x)^{v(x)}$ 的对数求导法。(a) $y = x^{\sin x}$ 的导数。(b) 在 $\pi/2$ 处求值。(c) $z = (x^{2} + 1)^{x^{2}}$。(d) 一般公式与一致性验证。

Answers:答案：  (b) $\dfrac{dy}{dx}\Big|_{\pi/2} \approx 1.00$  ·  (c) $\dfrac{dz}{dx} = 2 x\,(x^{2} + 1)^{x^{2}}\left[\ln(x^{2} + 1) + \dfrac{x^{2}}{x^{2} + 1}\right]$

(a) Logarithmic differentiation of $y = x^{\sin x}$ M1·A1·A1·A1

Take $\ln$ of both sides (valid since $x > 0$): $\ln y = \sin x \cdot \ln x$. Differentiate implicitly w.r.t. $x$, using product rule on the right: $$ \frac{1}{y} \cdot \frac{dy}{dx} \;=\; \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}. $$ Multiply by $y$ and reinstate $y = x^{\sin x}$: $$ \frac{dy}{dx} \;=\; x^{\sin x}\left(\cos x \,\ln x + \frac{\sin x}{x}\right). \quad \text{AG} $$

(b) Evaluate at $x = \pi/2$ M1·A1

At $x = \pi/2$: $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$, $\ln(\pi/2) \approx 0.4516$, $x^{\sin x} = (\pi/2)^{1} = \pi/2 \approx 1.5708$. Formula: $$ \frac{dy}{dx}\bigg|_{\pi/2} = \frac{\pi}{2}\left(0 \cdot \ln(\pi/2) + \frac{1}{\pi/2}\right) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1. $$ GDC nDeriv of $x^{\sin x}$ at $\pi/2$: $\approx 1.00$. $\checkmark$ (3 s.f.).

(c) $z = (x^{2} + 1)^{x^{2}}$ M1·A1·A1·A1

$\ln z = x^{2} \cdot \ln(x^{2} + 1)$. Differentiate by product + chain rule: $$ \frac{1}{z}\,\frac{dz}{dx} \;=\; 2 x \cdot \ln(x^{2} + 1) + x^{2} \cdot \frac{2 x}{x^{2} + 1} \;=\; 2 x\left[\ln(x^{2} + 1) + \frac{x^{2}}{x^{2} + 1}\right]. $$ Hence $$ \frac{dz}{dx} \;=\; 2 x\,(x^{2} + 1)^{x^{2}}\left[\ln(x^{2} + 1) + \frac{x^{2}}{x^{2} + 1}\right]. $$

(d) General formula M1·A1·A1·A1·R1

Let $w = u(x)^{v(x)}$ with $u(x) > 0$. Take $\ln$: $\ln w = v(x)\,\ln u(x)$. Differentiate implicitly using product + chain rule: $$ \frac{1}{w}\,\frac{dw}{dx} \;=\; v'(x)\,\ln u(x) + v(x) \cdot \frac{u'(x)}{u(x)}. $$ Multiply by $w = u(x)^{v(x)}$: $$ \frac{dw}{dx} \;=\; u(x)^{v(x)}\left[v'(x)\,\ln u(x) + \frac{v(x)\,u'(x)}{u(x)}\right]. \quad \text{AG} $$ Consistency with (a). Take $u(x) = x$, $v(x) = \sin x$, so $u' = 1$, $v' = \cos x$, $\ln u = \ln x$. Substitute: $$ \frac{dw}{dx} = x^{\sin x}\!\left[\cos x \,\ln x + \frac{\sin x \cdot 1}{x}\right] = x^{\sin x}\!\left(\cos x \,\ln x + \frac{\sin x}{x}\right). \;\checkmark $$ Consistency with (c). Take $u(x) = x^{2} + 1$, $v(x) = x^{2}$, so $u' = 2 x$, $v' = 2 x$, $\ln u = \ln(x^{2} + 1)$. Substitute: $$ \frac{dw}{dx} = (x^{2}+1)^{x^{2}}\!\left[2 x \,\ln(x^{2}+1) + \frac{x^{2} \cdot 2 x}{x^{2} + 1}\right] = 2 x\,(x^{2}+1)^{x^{2}}\!\left[\ln(x^{2}+1) + \frac{x^{2}}{x^{2}+1}\right]. \;\checkmark $$
The two-term split decodes the formula. The general derivative $\dfrac{dw}{dx} = w\left[v'\ln u + \dfrac{v u'}{u}\right]$ has a clean interpretation: the first term $v'\ln u$ is what you'd get treating $u$ as a constant (so $w = u^{v}$ behaves like $a^{v}$ with derivative $a^{v}\ln a$); the second term $v u'/u$ is what you'd get treating $v$ as a constant (so $w = u^{v}$ behaves like $u^{n}$ with derivative $n u^{n - 1} u' = v u^{v - 1} u' = w \cdot v u'/u$). Total derivative = "constant-$u$" piece + "constant-$v$" piece. This decomposition is exactly the chain rule applied to $w(u, v)$ viewed as a function of two variables — an early glimpse of multivariable calculus. Pattern-match in exams: whenever both base and exponent depend on $x$, write $\ln$ first, then read off two natural pieces; you will not need to memorise the final formula.

(a) $y = x^{\sin x}$ 的对数求导 M1·A1·A1·A1

两边取 $\ln$（$x > 0$ 合法）：$\ln y = \sin x \cdot \ln x$。两边对 $x$ 隐式求导，右边用乘积法则： $$ \frac{1}{y} \cdot \frac{dy}{dx} \;=\; \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}. $$ 两边乘 $y$，并代回 $y = x^{\sin x}$： $$ \frac{dy}{dx} \;=\; x^{\sin x}\left(\cos x \,\ln x + \frac{\sin x}{x}\right). \quad \text{AG} $$

(b) $x = \pi/2$ 处求值 M1·A1

在 $x = \pi/2$：$\sin(\pi/2) = 1$、$\cos(\pi/2) = 0$、$\ln(\pi/2) \approx 0.4516$、$x^{\sin x} = (\pi/2)^{1} = \pi/2 \approx 1.5708$。代入公式： $$ \frac{dy}{dx}\bigg|_{\pi/2} = \frac{\pi}{2}\left(0 \cdot \ln(\pi/2) + \frac{1}{\pi/2}\right) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1. $$ GDC nDeriv 对 $x^{\sin x}$ 在 $\pi/2$ 处给 $\approx 1.00$。$\checkmark$（$3$ 位有效数字）。

(c) $z = (x^{2} + 1)^{x^{2}}$ M1·A1·A1·A1

$\ln z = x^{2} \cdot \ln(x^{2} + 1)$。用乘积 + 链式法则求导： $$ \frac{1}{z}\,\frac{dz}{dx} \;=\; 2 x \cdot \ln(x^{2} + 1) + x^{2} \cdot \frac{2 x}{x^{2} + 1} \;=\; 2 x\left[\ln(x^{2} + 1) + \frac{x^{2}}{x^{2} + 1}\right]. $$ 故 $$ \frac{dz}{dx} \;=\; 2 x\,(x^{2} + 1)^{x^{2}}\left[\ln(x^{2} + 1) + \frac{x^{2}}{x^{2} + 1}\right]. $$

(d) 一般公式 M1·A1·A1·A1·R1

设 $w = u(x)^{v(x)}$，$u(x) > 0$。取 $\ln$：$\ln w = v(x)\,\ln u(x)$。用乘积 + 链式法则隐式求导： $$ \frac{1}{w}\,\frac{dw}{dx} \;=\; v'(x)\,\ln u(x) + v(x) \cdot \frac{u'(x)}{u(x)}. $$ 两边乘 $w = u(x)^{v(x)}$： $$ \frac{dw}{dx} \;=\; u(x)^{v(x)}\left[v'(x)\,\ln u(x) + \frac{v(x)\,u'(x)}{u(x)}\right]. \quad \text{AG} $$ 与 (a) 一致。取 $u(x) = x$、$v(x) = \sin x$，$u' = 1$、$v' = \cos x$、$\ln u = \ln x$，代入： $$ \frac{dw}{dx} = x^{\sin x}\!\left[\cos x \,\ln x + \frac{\sin x \cdot 1}{x}\right] = x^{\sin x}\!\left(\cos x \,\ln x + \frac{\sin x}{x}\right). \;\checkmark $$ 与 (c) 一致。取 $u(x) = x^{2} + 1$、$v(x) = x^{2}$，$u' = 2 x$、$v' = 2 x$、$\ln u = \ln(x^{2} + 1)$，代入： $$ \frac{dw}{dx} = (x^{2}+1)^{x^{2}}\!\left[2 x \,\ln(x^{2}+1) + \frac{x^{2} \cdot 2 x}{x^{2} + 1}\right] = 2 x\,(x^{2}+1)^{x^{2}}\!\left[\ln(x^{2}+1) + \frac{x^{2}}{x^{2}+1}\right]. \;\checkmark $$
两项拆解破解公式。一般导数 $\dfrac{dw}{dx} = w\left[v'\ln u + \dfrac{v u'}{u}\right]$ 有清晰意义：第一项 $v'\ln u$ 来自把 $u$ 视为常数（此时 $w = u^{v}$ 形如 $a^{v}$，导数 $a^{v}\ln a$）；第二项 $v u'/u$ 来自把 $v$ 视为常数（此时 $w = u^{v}$ 形如 $u^{n}$，导数 $n u^{n - 1} u' = v u^{v - 1} u' = w \cdot v u'/u$）。总导数 $=$ "$u$ 为常数"贡献 $+$ "$v$ 为常数"贡献。这正是把 $w(u, v)$ 视为两元函数后的链式法则——多元微积分的预演。考场上凡是底数与指数都依赖 $x$ 的，先写 $\ln$，再读两个自然部分；不必背最终公式。