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Chapter 3 · Solutions第三章 · 解析

Composite, Implicit & Inverse Functions · Solutions复合、隐函数与反函数 · 解析

Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析

EASYMEDIUMHARD

Topics 3.1-3.6考点 3.1 至 3.6AB

PART ITopics 3.1-3.6考点 3.1 至 3.6

Multiple Choice Solutions选择题解析

Q1EASY 3.1 Chain Rule3.1 链式法则No Calculator[2 marks]

If $f(x)=(3x^{2}+1)^{4}$, then $f'(x)=$若 $f(x)=(3x^{2}+1)^{4}$，则 $f'(x)=$

Answer:答案： (B) $24x(3x^{2}+1)^{3}$

Differentiate the outer power, then the inner function先对外层幂函数求导，再对内层函数求导 M1·A1

Treat $f$ as $u^{4}$ with inner function $u=3x^{2}+1$: the chain rule gives $f'(x)=4u^{3}\cdot u'=4(3x^{2}+1)^{3}\cdot 6x$. (M1)将 $f$ 视为 $u^{4}$，其中内函数 $u=3x^{2}+1$：由链式法则得 $f'(x)=4u^{3}\cdot u'=4(3x^{2}+1)^{3}\cdot 6x$。(M1)

Simplify the constant: $4\cdot 6x=24x$, so $f'(x)=24x(3x^{2}+1)^{3}$. (A1)化简常数：$4\cdot 6x=24x$，故 $f'(x)=24x(3x^{2}+1)^{3}$。(A1)

Insight.要点。 Option (A) is the classic chain-rule trap: differentiating only the outer power and forgetting to multiply by the inner derivative $u'=6x$. Always name the inner function first, then apply "outer prime times inner prime" as two separate steps.选项 (A) 是链式法则的典型陷阱：只对外层幂函数求导，却忘记乘以内层导数 $u'=6x$。应始终先明确内函数，再分两步应用"外层导数乘内层导数"。

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Q2EASY 3.1 Chain (Trig)3.1 链式法则（三角）No Calculator[2 marks]

$\dfrac{d}{dx}\bigl[\,\sin(4x)\,\bigr]=$

Answer:答案： (B) $4\cos(4x)$

Outer derivative of $\sin$ times inner derivative$\sin$ 的外层导数乘以内层导数 M1·A1

With inner function $u=4x$, $\dfrac{d}{dx}[\sin u]=\cos u\cdot u'=\cos(4x)\cdot 4$. (M1)令内函数 $u=4x$，$\dfrac{d}{dx}[\sin u]=\cos u\cdot u'=\cos(4x)\cdot 4$。(M1)

So $\dfrac{d}{dx}[\sin(4x)]=4\cos(4x)$. (A1)故 $\dfrac{d}{dx}[\sin(4x)]=4\cos(4x)$。(A1)

Insight.要点。 The coefficient inside a trig argument always reappears as a multiplier out front after differentiating: $\dfrac{d}{dx}[\sin(kx)]=k\cos(kx)$ and $\dfrac{d}{dx}[\cos(kx)]=-k\sin(kx)$ for any constant $k$ are worth having memorized as instant patterns.三角函数自变量中的系数，求导后总会以乘数形式重新出现在最前面：对任意常数 $k$，$\dfrac{d}{dx}[\sin(kx)]=k\cos(kx)$ 且 $\dfrac{d}{dx}[\cos(kx)]=-k\sin(kx)$，这两个模式值得直接记住。

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Q3EASY 3.1 Chain (Exp)3.1 链式法则（指数）No Calculator[2 marks]

$\dfrac{d}{dx}\bigl[\,e^{\,x^{2}}\,\bigr]=$

Answer:答案： (B) $2x\,e^{\,x^{2}}$

$e^{u}$ reproduces itself, times $u'$$e^{u}$ 求导后自我复现，再乘以 $u'$ M1·A1

With inner function $u=x^{2}$, $\dfrac{d}{dx}[e^{u}]=e^{u}\cdot u'=e^{x^{2}}\cdot 2x$. (M1)令内函数 $u=x^{2}$，$\dfrac{d}{dx}[e^{u}]=e^{u}\cdot u'=e^{x^{2}}\cdot 2x$。(M1)

So $\dfrac{d}{dx}[e^{x^{2}}]=2x\,e^{x^{2}}$. (A1)故 $\dfrac{d}{dx}[e^{x^{2}}]=2x\,e^{x^{2}}$。(A1)

Insight.要点。 $e^{u}$ is the one elementary function whose derivative is itself, so the entire chain-rule computation reduces to finding $u'$ and reattaching it as a multiplier. Option (C) is the trap of applying the power rule to the exponent instead of leaving $e^{x^2}$ intact.$e^{u}$ 是唯一一个求导后等于自身的初等函数，因此整个链式法则计算最终归结为求出 $u'$ 并将其作为乘数重新附加。选项 (C) 的陷阱在于对指数部分套用幂法则，而不是保留 $e^{x^2}$ 本身不变。

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Q4MEDIUM 3.1 Chain (Log)3.1 链式法则（对数）No Calculator[2 marks]

$\dfrac{d}{dx}\bigl[\ln(x^{2}+3x)\bigr]=$

Answer:答案： (B) $\dfrac{2x+3}{x^{2}+3x}$

$\ln u$ differentiates to $u'/u$$\ln u$ 求导为 $u'/u$ M1·A1

With inner function $u=x^{2}+3x$, $\dfrac{d}{dx}[\ln u]=\dfrac{u'}{u}$, and $u'=2x+3$. (M1)令内函数 $u=x^{2}+3x$，$\dfrac{d}{dx}[\ln u]=\dfrac{u'}{u}$，且 $u'=2x+3$。(M1)

So $\dfrac{d}{dx}[\ln(x^{2}+3x)]=\dfrac{2x+3}{x^{2}+3x}$. (A1)故 $\dfrac{d}{dx}[\ln(x^{2}+3x)]=\dfrac{2x+3}{x^{2}+3x}$。(A1)

Insight.要点。 $\dfrac{d}{dx}[\ln u]=\dfrac{u'}{u}$ is the pattern to reach for immediately whenever a logarithm's argument is anything other than plain $x$; option (C) confuses this quotient with dividing the derivative by $\ln u$ itself, which is not a rule.只要对数的自变量不是单纯的 $x$，就应立即联想到模式 $\dfrac{d}{dx}[\ln u]=\dfrac{u'}{u}$；选项 (C) 将此商式误认为是导数除以 $\ln u$ 本身，这并非有效法则。

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Q5MEDIUM 3.1 Nested Chain3.1 嵌套链式法则No Calculator[2 marks]

If $y=\sin^{2}(3x)$, then $\dfrac{dy}{dx}=$若 $y=\sin^{2}(3x)$，则 $\dfrac{dy}{dx}=$

Answer:答案： (B) $6\sin(3x)\cos(3x)$

Peel two layers: the square, then the sine剥离两层结构：先平方，再正弦 M1·A1

Read $y=[\sin(3x)]^{2}$ as a square of an inner function $u=\sin(3x)$: $\dfrac{dy}{dx}=2u\cdot u'=2\sin(3x)\cdot\dfrac{d}{dx}[\sin(3x)]$. (M1)将 $y=[\sin(3x)]^{2}$ 视为内函数 $u=\sin(3x)$ 的平方：$\dfrac{dy}{dx}=2u\cdot u'=2\sin(3x)\cdot\dfrac{d}{dx}[\sin(3x)]$。(M1)

The inner derivative itself needs the chain rule: $\dfrac{d}{dx}[\sin(3x)]=3\cos(3x)$, so $\dfrac{dy}{dx}=2\sin(3x)\cdot 3\cos(3x)=6\sin(3x)\cos(3x)$. (A1)内层导数本身仍需链式法则：$\dfrac{d}{dx}[\sin(3x)]=3\cos(3x)$，故 $\dfrac{dy}{dx}=2\sin(3x)\cdot 3\cos(3x)=6\sin(3x)\cos(3x)$。(A1)

Insight.要点。 A composition of three functions ("square of sine of $3x$") needs the chain rule applied twice, once per layer, from the outside in. Option (D) stops after only the outer layer and forgets the factor of $3$ from $\dfrac{d}{dx}[3x]$: always count the layers before differentiating.三层复合函数（"$3x$ 的正弦的平方"）需要从外到内应用两次链式法则，每层一次。选项 (D) 只完成了外层就停止，遗漏了来自 $\dfrac{d}{dx}[3x]$ 的系数 $3$：求导前应先数清复合的层数。

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Q6MEDIUM 3.1 Chain (Table)3.1 链式法则（表格）No Calculator[2 marks]

Selected values of $f$ and $g$; find $h'(1)$ for $h(x)=f(g(x))$.$f$ 和 $g$ 的部分函数值；求 $h(x)=f(g(x))$ 的 $h'(1)$。

Answer:答案： (C) $2$

Read $g(1)$ and $g'(1)$, then look up $f'$ at that value读出 $g(1)$ 与 $g'(1)$，再查出对应的 $f'$ 值 M1·A1

By the chain rule, $h'(x)=f'(g(x))\cdot g'(x)$, so $h'(1)=f'(g(1))\cdot g'(1)$. From the table, $g(1)=3$ and $g'(1)=-2$. (M1)由链式法则，$h'(x)=f'(g(x))\cdot g'(x)$，故 $h'(1)=f'(g(1))\cdot g'(1)$。由表格，$g(1)=3$，$g'(1)=-2$。(M1)

Looking up $x=3$ in the table gives $f'(3)=-1$, so $h'(1)=f'(3)\cdot g'(1)=(-1)(-2)=2$. (A1)在表中查 $x=3$ 得 $f'(3)=-1$，故 $h'(1)=f'(3)\cdot g'(1)=(-1)(-2)=2$。(A1)

Insight.要点。 The most common table-chain-rule error is evaluating $f'$ at $x=1$ instead of at $g(1)=3$: the outer function's derivative must be evaluated at the inner function's output, not at the original input. Read the table twice, once for $g(1)$ and $g'(1)$, once more for $f'$ at that output.表格链式法则最常见的错误是在 $x=1$ 而非 $g(1)=3$ 处求 $f'$ 的值：外层函数的导数必须在内层函数的输出值处求值，而非原始输入处。应两次查表：先查 $g(1)$ 与 $g'(1)$，再在该输出值处查 $f'$。

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Q7MEDIUM 3.2 Implicit3.2 隐函数微分No Calculator[2 marks]

If $x^{2}+y^{2}=25$, then $\dfrac{dy}{dx}=$若 $x^{2}+y^{2}=25$，则 $\dfrac{dy}{dx}=$

Answer:答案： (A) $-\dfrac{x}{y}$

Differentiate every term with respect to $x$, treating $y$ as $y(x)$对每一项关于 $x$ 求导，将 $y$ 视为 $y(x)$ M1·A1

Differentiate both sides: $\dfrac{d}{dx}[x^{2}]+\dfrac{d}{dx}[y^{2}]=\dfrac{d}{dx}[25]$ gives $2x+2y\dfrac{dy}{dx}=0$, using the chain rule on $y^{2}$. (M1)两边求导：$\dfrac{d}{dx}[x^{2}]+\dfrac{d}{dx}[y^{2}]=\dfrac{d}{dx}[25]$，对 $y^{2}$ 用链式法则得 $2x+2y\dfrac{dy}{dx}=0$。(M1)

Solve for $\dfrac{dy}{dx}$: $2y\dfrac{dy}{dx}=-2x$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}$. (A1)解出 $\dfrac{dy}{dx}$：$2y\dfrac{dy}{dx}=-2x$，故 $\dfrac{dy}{dx}=-\dfrac{x}{y}$。(A1)

Insight.要点。 Every $y$-term picks up a factor of $\dfrac{dy}{dx}$ from the chain rule since $y$ is itself a function of $x$: this is the single idea behind all implicit differentiation. On a circle, the result $-x/y$ also matches the geometric fact that a radius and its tangent line are perpendicular.因 $y$ 本身是 $x$ 的函数，每个含 $y$ 的项经链式法则都会带上一个因子 $\dfrac{dy}{dx}$：这正是隐函数微分背后唯一的核心思想。对于圆而言，结果 $-x/y$ 也恰好对应半径与其切线互相垂直这一几何事实。

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Q8MEDIUM 3.2 Implicit (Mixed)3.2 隐函数（混合型）No Calculator[2 marks]

If $xy+y^{3}=4$, then $\dfrac{dy}{dx}=$若 $xy+y^{3}=4$，则 $\dfrac{dy}{dx}=$

Answer:答案： (A) $-\dfrac{y}{x+3y^{2}}$

Product rule on $xy$, chain rule on $y^{3}$对 $xy$ 用乘积法则，对 $y^{3}$ 用链式法则 M1·A1

Differentiate: $\dfrac{d}{dx}[xy]=y+x\dfrac{dy}{dx}$ (product rule) and $\dfrac{d}{dx}[y^{3}]=3y^{2}\dfrac{dy}{dx}$ (chain rule), so $y+x\dfrac{dy}{dx}+3y^{2}\dfrac{dy}{dx}=0$. (M1)求导：$\dfrac{d}{dx}[xy]=y+x\dfrac{dy}{dx}$（乘积法则），$\dfrac{d}{dx}[y^{3}]=3y^{2}\dfrac{dy}{dx}$（链式法则），故 $y+x\dfrac{dy}{dx}+3y^{2}\dfrac{dy}{dx}=0$。(M1)

Collect all $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(x+3y^{2})=-y$, so $\dfrac{dy}{dx}=-\dfrac{y}{x+3y^{2}}$. (A1)将所有 $\dfrac{dy}{dx}$ 项归并：$\dfrac{dy}{dx}(x+3y^{2})=-y$，故 $\dfrac{dy}{dx}=-\dfrac{y}{x+3y^{2}}$。(A1)

Insight.要点。 A mixed term like $xy$ always needs the product rule, not just the chain rule, because both factors depend on $x$: this is the step that trips up option-(C)-style errors, which drop the $y$ term from differentiating $xy$ entirely. Isolate every $\dfrac{dy}{dx}$ term on one side before factoring.$xy$ 这类混合项总是需要用乘积法则，而非仅用链式法则，因为两个因子都依赖于 $x$：这正是选项 (C) 一类错误的根源，即在对 $xy$ 求导时完全遗漏了 $y$ 这一项。应先将所有 $\dfrac{dy}{dx}$ 项归并到一侧，再提取因式。

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Q9HARD 3.2 Implicit Tangent3.2 隐函数切线No Calculator[3 marks]

For the curve $x^{2}+xy+y^{2}=7$, the slope of the tangent line at $(1,2)$ is曲线 $x^{2}+xy+y^{2}=7$ 在点 $(1,2)$ 处切线的斜率为

Answer:答案： (A) $-\dfrac{4}{5}$

Differentiate the mixed term with the product rule对混合项用乘积法则求导 M1·M1·A1

Differentiate every term: $2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$, where the middle term uses the product rule on $xy$. (M1)对每一项求导：$2x+\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$，其中中间项对 $xy$ 用了乘积法则。(M1)

Collect $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(x+2y)=-(2x+y)$, so $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$. (M1)归并 $\dfrac{dy}{dx}$ 项：$\dfrac{dy}{dx}(x+2y)=-(2x+y)$，故 $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$。(M1)

Substitute $(x,y)=(1,2)$: $\dfrac{dy}{dx}=-\dfrac{2(1)+2}{1+2(2)}=-\dfrac{4}{5}$. (A1)代入 $(x,y)=(1,2)$：$\dfrac{dy}{dx}=-\dfrac{2(1)+2}{1+2(2)}=-\dfrac{4}{5}$。(A1)

Insight.要点。 Always finish the algebra, an explicit formula for $\dfrac{dy}{dx}$ in terms of $x$ and $y$, before plugging in numbers: substituting too early into a half-differentiated equation is a common source of arithmetic slips on this exact question type.务必先完成代数运算，得到用 $x$、$y$ 表示的 $\dfrac{dy}{dx}$ 显式表达式，再代入具体数值：过早地将数值代入尚未求导完毕的方程，正是此类题目算术出错的常见根源。

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Q10EASY 3.3 Inverse Derivative3.3 反函数导数No Calculator[3 marks]

If $f(x)=x^{3}+x+1$, then $(f^{-1})'(1)=$若 $f(x)=x^{3}+x+1$，则 $(f^{-1})'(1)=$

Answer:答案： (C) $1$

Find the input that $f$ sends to $1$求出 $f$ 映射到 $1$ 的输入值 M1·M1·A1

The inverse-derivative formula $(f^{-1})'(a)=\dfrac{1}{f'(f^{-1}(a))}$ needs $f^{-1}(1)$ first: solve $x^{3}+x+1=1$, i.e. $x^{3}+x=0$, i.e. $x(x^{2}+1)=0$. Since $x^{2}+1>0$ always, the only real root is $x=0$, so $f^{-1}(1)=0$. (M1)反函数导数公式 $(f^{-1})'(a)=\dfrac{1}{f'(f^{-1}(a))}$ 需要先求出 $f^{-1}(1)$：解 $x^{3}+x+1=1$，即 $x^{3}+x=0$，即 $x(x^{2}+1)=0$。因 $x^{2}+1>0$ 恒成立，唯一实根为 $x=0$，故 $f^{-1}(1)=0$。(M1)

Compute $f'(x)=3x^{2}+1$ and evaluate at $x=0$: $f'(0)=1$. (M1)求 $f'(x)=3x^{2}+1$，并在 $x=0$ 处求值：$f'(0)=1$。(M1)

So $(f^{-1})'(1)=\dfrac{1}{f'(0)}=\dfrac{1}{1}=1$. (A1)故 $(f^{-1})'(1)=\dfrac{1}{f'(0)}=\dfrac{1}{1}=1$。(A1)

Insight.要点。 The inverse-derivative formula never asks you to actually find a formula for $f^{-1}(x)$ (which may be impossible in closed form for a cubic like this one): it only needs the single input value $f^{-1}(a)$, found by solving $f(x)=a$, then evaluating $f'$ (not $(f^{-1})'$) at that value.反函数导数公式从不要求真正求出 $f^{-1}(x)$ 的解析表达式（对于这样的三次函数，其反函数往往无法用初等形式表示）：它只需要单个输入值 $f^{-1}(a)$，通过解方程 $f(x)=a$ 得到，再在该值处求 $f'$（而非 $(f^{-1})'$）。

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Q11MEDIUM 3.3 Inverse (Table)3.3 反函数（表格）No Calculator[2 marks]

Let $f$ be differentiable and one-to-one, with selected values in the table. Find $(f^{-1})'(7)$.设 $f$ 可微且为一一映射，部分函数值如表所示，求 $(f^{-1})'(7)$。

Answer:答案： (B) $\dfrac{1}{5}$

Find which $x$ satisfies $f(x)=7$, then read off $f'$ there求出满足 $f(x)=7$ 的 $x$，再查出该处的 $f'$ M1·A1

From the table, $f(2)=7$, so $f^{-1}(7)=2$. (M1)由表格，$f(2)=7$，故 $f^{-1}(7)=2$。(M1)

By the inverse-derivative formula, $(f^{-1})'(7)=\dfrac{1}{f'(f^{-1}(7))}=\dfrac{1}{f'(2)}=\dfrac{1}{5}$. (A1)由反函数导数公式，$(f^{-1})'(7)=\dfrac{1}{f'(f^{-1}(7))}=\dfrac{1}{f'(2)}=\dfrac{1}{5}$。(A1)

Insight.要点。 Reading the table in the wrong direction, using $f'(7)$ instead of $f'(f^{-1}(7))=f'(2)$, is the single most common error on this question type. Always locate $7$ in the $f(x)$ row first, then move up to the matching $x$, and only then read $f'$ off that same column.从错误方向读表，即误用 $f'(7)$ 而非 $f'(f^{-1}(7))=f'(2)$，是此类题目中最常见的错误。应先在 $f(x)$ 行中找到 $7$，再上移至对应的 $x$，最后在该列读出 $f'$。

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Q12EASY 3.4 Inverse Trig3.4 反三角函数No Calculator[2 marks]

$\dfrac{d}{dx}\bigl[\arctan x\bigr]=$

Answer:答案： (A) $\dfrac{1}{1+x^{2}}$

Apply the memorized inverse-trig derivative套用已记忆的反三角函数导数公式 M1·A1

The standard derivative formula is $\dfrac{d}{dx}[\arctan x]=\dfrac{1}{1+x^{2}}$. (M1)标准导数公式为 $\dfrac{d}{dx}[\arctan x]=\dfrac{1}{1+x^{2}}$。(M1)

This matches option (A) directly, with plain $x$ as the argument (not a composed inner function), so no further chain-rule multiplier is needed. (A1)该式直接对应选项 (A)，因自变量为单纯的 $x$（并非复合的内函数），无需再乘以链式法则的乘子。(A1)

Insight.要点。 $\arcsin$, $\arccos$, and $\arctan$ each have a derivative formula worth memorizing outright: $\dfrac{1}{\sqrt{1-x^{2}}}$, $\dfrac{-1}{\sqrt{1-x^{2}}}$, and $\dfrac{1}{1+x^{2}}$ respectively. Option (D) is the trap of confusing $\arctan x$ with $\tan x$, whose derivative is $\sec^{2}x$: the two are inverse functions of each other, with very different derivatives.$\arcsin$、$\arccos$、$\arctan$ 各自的导数公式都值得直接背诵：分别为 $\dfrac{1}{\sqrt{1-x^{2}}}$、$\dfrac{-1}{\sqrt{1-x^{2}}}$、$\dfrac{1}{1+x^{2}}$。选项 (D) 的陷阱在于将 $\arctan x$ 与 $\tan x$ 混淆，后者的导数为 $\sec^{2}x$：二者互为反函数，导数截然不同。

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Q13MEDIUM 3.4 Inverse Trig Chain3.4 反三角链式法则No Calculator[2 marks]

$\dfrac{d}{dx}\bigl[\arcsin(2x)\bigr]=$

Answer:答案： (B) $\dfrac{2}{\sqrt{1-4x^{2}}}$

Substitute $u=2x$ into the arcsin formula, then multiply by $u'$将 $u=2x$ 代入反正弦公式，再乘以 $u'$ M1·A1

With inner function $u=2x$, $\dfrac{d}{dx}[\arcsin u]=\dfrac{u'}{\sqrt{1-u^{2}}}=\dfrac{2}{\sqrt{1-(2x)^{2}}}$. (M1)令内函数 $u=2x$，$\dfrac{d}{dx}[\arcsin u]=\dfrac{u'}{\sqrt{1-u^{2}}}=\dfrac{2}{\sqrt{1-(2x)^{2}}}$。(M1)

Simplify $(2x)^{2}=4x^{2}$: $\dfrac{d}{dx}[\arcsin(2x)]=\dfrac{2}{\sqrt{1-4x^{2}}}$. (A1)化简 $(2x)^{2}=4x^{2}$：$\dfrac{d}{dx}[\arcsin(2x)]=\dfrac{2}{\sqrt{1-4x^{2}}}$。(A1)

Insight.要点。 Option (A) drops the chain-rule multiplier $u'=2$, and option (C) forgets to square the $2$ inside the radical: the inverse-trig formulas are chain-rule formulas, so both the outer derivative's own structure ($\sqrt{1-u^2}$, with $u$ squared) and the inner derivative $u'$ must be tracked separately.选项 (A) 遗漏了链式法则的乘数 $u'=2$，选项 (C) 忘记对根号内的 $2$ 取平方：反三角函数公式本身即为链式法则公式，因此外层导数自身的结构（$\sqrt{1-u^2}$ 中 $u$ 需取平方）与内层导数 $u'$ 须分别单独跟踪。

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Q14MEDIUM 3.5 Higher-Order Implicit3.5 高阶隐函数导数No Calculator[3 marks]

If $x^{2}+y^{2}=4$, then $\dfrac{d^{2}y}{dx^{2}}=$若 $x^{2}+y^{2}=4$，则 $\dfrac{d^{2}y}{dx^{2}}=$

Answer:答案： (B) $-\dfrac{4}{y^{3}}$

Differentiate implicitly once for $y'$先隐函数微分一次求 $y'$ M1·M1·A1

Differentiating $x^{2}+y^{2}=4$ gives $2x+2yy'=0$, so $x+yy'=0$ and $y'=-\dfrac{x}{y}$. (M1)对 $x^{2}+y^{2}=4$ 求导得 $2x+2yy'=0$，故 $x+yy'=0$，$y'=-\dfrac{x}{y}$。(M1)

Differentiate $x+yy'=0$ again with the product rule on $yy'$: $1+(y')^{2}+yy''=0$, so $y''=-\dfrac{1+(y')^{2}}{y}$. Substituting $(y')^{2}=\dfrac{x^{2}}{y^{2}}$ gives $y''=-\dfrac{y^{2}+x^{2}}{y^{3}}=-\dfrac{x^{2}+y^{2}}{y^{3}}$. (M1)再对 $x+yy'=0$ 求导，对 $yy'$ 用乘积法则：$1+(y')^{2}+yy''=0$，故 $y''=-\dfrac{1+(y')^{2}}{y}$。代入 $(y')^{2}=\dfrac{x^{2}}{y^{2}}$ 得 $y''=-\dfrac{y^{2}+x^{2}}{y^{3}}=-\dfrac{x^{2}+y^{2}}{y^{3}}$。(M1)

Substitute the curve's own equation $x^{2}+y^{2}=4$ to simplify: $y''=-\dfrac{4}{y^{3}}$. (A1)代入曲线本身的方程 $x^{2}+y^{2}=4$ 化简：$y''=-\dfrac{4}{y^{3}}$。(A1)

Insight.要点。 Note that option (D), $-\dfrac{x^{2}+y^{2}}{y^{3}}$, is numerically identical to (B) at every point of this specific curve, since $x^{2}+y^{2}=4$ there by definition: (D) is the general unsimplified formula (valid for any circle $x^{2}+y^{2}=r^{2}$, giving $y''=-r^{2}/y^{3}$), while (B) is that same result after substituting this curve's own constant. Always finish by using the original equation to eliminate leftover $x$'s when the question gives you a specific curve.注意选项 (D)，即 $-\dfrac{x^{2}+y^{2}}{y^{3}}$，在这条特定曲线上的每一点都与 (B) 在数值上完全相同，因为该曲线上恒有 $x^{2}+y^{2}=4$：(D) 是尚未化简的一般公式（对任意圆 $x^{2}+y^{2}=r^{2}$ 均成立，给出 $y''=-r^{2}/y^{3}$），而 (B) 正是代入本曲线自身常数后的结果。当题目给出具体曲线时，应最终利用原方程消去残留的 $x$。

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Q15MEDIUM 4.4 Related Rates4.4 相关变化率Preview of Unit 4No Calculator[2 marks]

A circle's radius grows at $3$ cm/s. At $r=5$, find the rate of change of its area.某圆的半径以 $3$ cm/s 的速率增大，求当 $r=5$ 时其面积的变化率。

Answer:答案： (D) $30\pi$ cm²/s

Differentiate $A=\pi r^{2}$ implicitly with respect to time对 $A=\pi r^{2}$ 关于时间隐函数微分 M1·A1

Since $A=\pi r^{2}$ and both $A$ and $r$ depend on time $t$, the chain rule gives $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$. (M1)因 $A=\pi r^{2}$，且 $A$、$r$ 均随时间 $t$ 变化，由链式法则得 $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$。(M1)

Substitute $r=5$ and $\dfrac{dr}{dt}=3$: $\dfrac{dA}{dt}=2\pi(5)(3)=30\pi$ cm²/s. (A1)代入 $r=5$ 与 $\dfrac{dr}{dt}=3$：$\dfrac{dA}{dt}=2\pi(5)(3)=30\pi$ cm²/s。(A1)

Insight.要点。 Related rates is implicit differentiation with respect to $t$ instead of $x$: every variable that changes over time picks up its own $d(\cdot)/dt$ factor via the chain rule, exactly as $y$ picked up $dy/dx$ earlier in this unit. Always differentiate the general formula first, then substitute the specific instant.相关变化率就是对 $t$（而非 $x$）进行的隐函数微分：每个随时间变化的变量，都通过链式法则带上各自的 $d(\cdot)/dt$ 因子，正如本单元前面 $y$ 带上 $dy/dx$ 一样。应先对一般公式求导，再代入具体时刻的数值。

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Q16HARD 4.4 Related Rates (Ladder)4.4 相关变化率（梯子问题）Preview of Unit 4Calculator[3 marks]

A $13$-ft ladder slides down a wall. When the base is $5$ ft from the wall and moving at $2$ ft/s away, find the rate the top is moving.一把 $13$ 英尺长的梯子靠墙放置并向下滑动，求当底部距墙 $5$ 英尺且以 $2$ ft/s 的速率向外移动时梯子顶端的移动速率。

Answer:答案： (B) $-0.83$ ft/s

Set up the Pythagorean constraint, then find the missing side建立勾股定理约束，再求出缺失边长 M1·M1·A1

Let $x$ be the base's distance from the wall and $y$ the height of the top: $x^{2}+y^{2}=13^{2}=169$. At the instant $x=5$: $y=\sqrt{169-25}=\sqrt{144}=12$. (M1)设 $x$ 为底部距墙的距离，$y$ 为顶端的高度：$x^{2}+y^{2}=13^{2}=169$。当 $x=5$ 时：$y=\sqrt{169-25}=\sqrt{144}=12$。(M1)

Differentiate with respect to $t$: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$, so $\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}$. (M1)关于 $t$ 求导：$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$，故 $\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}$。(M1)

Substitute $x=5,\ y=12,\ \dfrac{dx}{dt}=2$: $\dfrac{dy}{dt}=-\dfrac{5}{12}(2)=-\dfrac{5}{6}\approx -0.83$ ft/s. (A1)代入 $x=5,\ y=12,\ \dfrac{dx}{dt}=2$：$\dfrac{dy}{dt}=-\dfrac{5}{12}(2)=-\dfrac{5}{6}\approx -0.83$ ft/s。(A1)

Insight.要点。 The negative sign is meaningful, not a mistake: $y$ (the height of the top) is decreasing as the ladder slides, while $x$ (the base's distance) increases, so their rates must have opposite signs, consistent with $x^{2}+y^{2}$ staying constant. Always find the missing side length with the constraint equation itself, at that specific instant, before differentiating rates.负号是有意义的，而非错误：随着梯子下滑，$y$（顶端高度）在减小，而 $x$（底部距离）在增大，故二者的变化率符号必然相反，这与 $x^{2}+y^{2}$ 保持不变相一致。求导前，应先用约束方程本身在该特定时刻求出缺失的边长。

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Q17HARD 3.2 Horizontal Tangent3.2 水平切线No Calculator[3 marks]

The curve $x^{2}-xy+y^{2}=3$ has a horizontal tangent at points where曲线 $x^{2}-xy+y^{2}=3$ 上水平切线的切点满足

Answer:答案： (B) $y=2x$

Differentiate implicitly, isolating $\dfrac{dy}{dx}$隐函数求导，解出 $\dfrac{dy}{dx}$ M1·M1·A1

Differentiate every term: $2x-\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$, using the product rule on $xy$. (M1)对每一项求导：$2x-\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0$，其中对 $xy$ 用了乘积法则。(M1)

Collect $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(2y-x)=y-2x$, so $\dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}$. (M1)归并 $\dfrac{dy}{dx}$ 项：$\dfrac{dy}{dx}(2y-x)=y-2x$，故 $\dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}$。(M1)

A horizontal tangent needs $\dfrac{dy}{dx}=0$, which happens exactly when the numerator vanishes (and the denominator does not): $y-2x=0$, i.e. $y=2x$. (A1)水平切线要求 $\dfrac{dy}{dx}=0$，这恰好在分子为零（且分母不为零）时成立：$y-2x=0$，即 $y=2x$。(A1)

Insight.要点。 A horizontal tangent is a numerator-equals-zero condition on $\dfrac{dy}{dx}$, exactly the mirror image of a vertical tangent, which is a denominator-equals-zero condition. Always solve for $\dfrac{dy}{dx}$ as a single fraction first, since the numerator and denominator conditions answer two different geometric questions.水平切线对应 $\dfrac{dy}{dx}$ 的分子为零这一条件，恰与竖直切线（分母为零的条件）互为镜像。应先将 $\dfrac{dy}{dx}$ 化简为单一分式，因为分子条件与分母条件回答的是两个不同的几何问题。

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Q18HARD 3.1 Chain + Product3.1 链式法则与乘积法则No Calculator[2 marks]

If $y=x^{2}\sin(\ln x)$, then $\dfrac{dy}{dx}=$若 $y=x^{2}\sin(\ln x)$，则 $\dfrac{dy}{dx}=$

Answer:答案： (D) both (A) and (B) are correct（A）和（B）均正确

Apply the product rule, then the chain rule on the $\sin(\ln x)$ factor先用乘积法则，再对 $\sin(\ln x)$ 因子用链式法则 M1·A1

With $u=x^{2}$ and $v=\sin(\ln x)$, the product rule gives $\dfrac{dy}{dx}=u'v+uv'=2x\sin(\ln x)+x^{2}\cdot\dfrac{d}{dx}[\sin(\ln x)]$. The chain rule gives $\dfrac{d}{dx}[\sin(\ln x)]=\cos(\ln x)\cdot\dfrac{1}{x}$, so $\dfrac{dy}{dx}=2x\sin(\ln x)+x^{2}\cos(\ln x)\cdot\dfrac{1}{x}$: this is exactly option (B). (M1)令 $u=x^{2}$，$v=\sin(\ln x)$，由乘积法则得 $\dfrac{dy}{dx}=u'v+uv'=2x\sin(\ln x)+x^{2}\cdot\dfrac{d}{dx}[\sin(\ln x)]$。由链式法则，$\dfrac{d}{dx}[\sin(\ln x)]=\cos(\ln x)\cdot\dfrac{1}{x}$，故 $\dfrac{dy}{dx}=2x\sin(\ln x)+x^{2}\cos(\ln x)\cdot\dfrac{1}{x}$：这正是选项 (B)。(M1)

Simplify $x^{2}\cdot\dfrac{1}{x}=x$: $\dfrac{dy}{dx}=2x\sin(\ln x)+x\cos(\ln x)$, which is exactly option (A). Since (A) is simply the fully simplified form of (B), both are correct expressions of the same derivative, so the answer is (D). (A1)化简 $x^{2}\cdot\dfrac{1}{x}=x$：$\dfrac{dy}{dx}=2x\sin(\ln x)+x\cos(\ln x)$，这正是选项 (A)。因 (A) 只是 (B) 完全化简后的形式，二者都是同一导数的正确表达，故答案为 (D)。(A1)

Insight.要点。 Multiple-choice traps are not always mathematically wrong answers, sometimes two options are algebraically identical and the question tests whether you can recognize a simplification step ($x^{2}\cdot\tfrac{1}{x}=x$) rather than a differentiation rule. Always fully simplify before comparing your result against the answer choices.选择题的陷阱并不总是数学上错误的答案，有时两个选项在代数上完全等价，题目考查的是能否识别一步化简（$x^{2}\cdot\tfrac{1}{x}=x$），而非求导法则本身。比对答案选项前，应先将结果完全化简。

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PART IIShow All Work展示完整解题过程

Free-Response Solutions自由解答题解析

FRQ 1EASY 3.1 Chain Rule3.1 链式法则No Calculator[6 marks]

Differentiate each function. Label inner and outer functions.对下列各函数求导，并标注内函数和外函数。

Answers:答案：  (a) $6(2x^{3}-5x+1)^{5}(6x^{2}-5)$  ·  (b) $-\dfrac{x}{\sqrt{9-x^{2}}}$  ·  (c) $-2e^{2x}\sin(e^{2x})$

(a) Outer function $u^{6}$, inner function $u=2x^{3}-5x+1$(a) 外函数 $u^{6}$，内函数 $u=2x^{3}-5x+1$ M1·A1

$u'=6x^{2}-5$, so $\dfrac{dy}{dx}=6u^{5}\cdot u'=6(2x^{3}-5x+1)^{5}(6x^{2}-5)$. (M1) The full expression is the final answer. (A1)$u'=6x^{2}-5$，故 $\dfrac{dy}{dx}=6u^{5}\cdot u'=6(2x^{3}-5x+1)^{5}(6x^{2}-5)$。(M1) 此完整表达式即为最终答案。(A1)

(b) Outer function $u^{1/2}$, inner function $u=9-x^{2}$(b) 外函数 $u^{1/2}$，内函数 $u=9-x^{2}$ M1·A1

Write $y=(9-x^{2})^{1/2}$ with $u'=-2x$: $\dfrac{dy}{dx}=\dfrac{1}{2}u^{-1/2}\cdot u'=\dfrac{1}{2}(9-x^{2})^{-1/2}(-2x)$. (M1)写 $y=(9-x^{2})^{1/2}$，$u'=-2x$：$\dfrac{dy}{dx}=\dfrac{1}{2}u^{-1/2}\cdot u'=\dfrac{1}{2}(9-x^{2})^{-1/2}(-2x)$。(M1)

Simplify: $\dfrac{dy}{dx}=-\dfrac{x}{\sqrt{9-x^{2}}}$. (A1)化简：$\dfrac{dy}{dx}=-\dfrac{x}{\sqrt{9-x^{2}}}$。(A1)

(c) Outer function $\cos u$, inner function $u=e^{2x}$(c) 外函数 $\cos u$，内函数 $u=e^{2x}$ M1·A1

$u'=2e^{2x}$ (itself a chain rule, on $e^{2x}$), so $\dfrac{dy}{dx}=-\sin u\cdot u'=-\sin(e^{2x})\cdot 2e^{2x}$. (M1) So $\dfrac{dy}{dx}=-2e^{2x}\sin(e^{2x})$. (A1)$u'=2e^{2x}$（本身也需链式法则，来自 $e^{2x}$），故 $\dfrac{dy}{dx}=-\sin u\cdot u'=-\sin(e^{2x})\cdot 2e^{2x}$。(M1) 故 $\dfrac{dy}{dx}=-2e^{2x}\sin(e^{2x})$。(A1)

Insight.要点。 Naming $u$ explicitly before differentiating, as instructed, turns every one of these into the same two-step recipe: differentiate the outer shape treating $u$ as a single variable, then multiply by $u'$. Part (c) shows the recipe can nest: the inner derivative $u'=2e^{2x}$ is itself a chain-rule computation.按要求在求导前显式命名 $u$，可将以上每一题都化为相同的两步流程：先把 $u$ 当作单一变量对外层结构求导，再乘以 $u'$。(c) 部分表明这一流程可以嵌套：内层导数 $u'=2e^{2x}$ 本身也需要一次链式法则计算。

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FRQ 2MEDIUM 3.2 Implicit Differentiation3.2 隐函数微分No Calculator[6 marks]

Consider the curve defined by $x^{2}+2xy+y^{3}=4$.考虑由 $x^{2}+2xy+y^{3}=4$ 所定义的曲线。

Answers:答案：  (a) $\dfrac{dy}{dx}=-\dfrac{2x+2y}{2x+3y^{2}}$  ·  (b) $(2,0)$ lies on the curve; slope在曲线上；斜率为 $=-1$  ·  (c) $y=-x+2$

(a) Product rule on $2xy$, chain rule on $y^{3}$(a) 对 $2xy$ 用乘积法则，对 $y^{3}$ 用链式法则 M1·M1·A1

Differentiate every term with respect to $x$: $2x+\dfrac{d}{dx}[2xy]+\dfrac{d}{dx}[y^{3}]=0$. (M1)对每一项关于 $x$ 求导：$2x+\dfrac{d}{dx}[2xy]+\dfrac{d}{dx}[y^{3}]=0$。(M1)

The product rule gives $\dfrac{d}{dx}[2xy]=2y+2x\dfrac{dy}{dx}$, and the chain rule gives $\dfrac{d}{dx}[y^{3}]=3y^{2}\dfrac{dy}{dx}$, so $2x+2y+2x\dfrac{dy}{dx}+3y^{2}\dfrac{dy}{dx}=0$. (M1)乘积法则给出 $\dfrac{d}{dx}[2xy]=2y+2x\dfrac{dy}{dx}$，链式法则给出 $\dfrac{d}{dx}[y^{3}]=3y^{2}\dfrac{dy}{dx}$，故 $2x+2y+2x\dfrac{dy}{dx}+3y^{2}\dfrac{dy}{dx}=0$。(M1)

Collect $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx}(2x+3y^{2})=-(2x+2y)$, so $\dfrac{dy}{dx}=-\dfrac{2x+2y}{2x+3y^{2}}$. (A1)归并 $\dfrac{dy}{dx}$ 项：$\dfrac{dy}{dx}(2x+3y^{2})=-(2x+2y)$，故 $\dfrac{dy}{dx}=-\dfrac{2x+2y}{2x+3y^{2}}$。(A1)

(b) Substitute the point into the curve equation, then into $\dfrac{dy}{dx}$(b) 将该点代入曲线方程，再代入 $\dfrac{dy}{dx}$ M1·A1

Check $(2,0)$: $2^{2}+2(2)(0)+0^{3}=4+0+0=4$, which matches the right-hand side, so $(2,0)$ lies on the curve. (M1)检验 $(2,0)$：$2^{2}+2(2)(0)+0^{3}=4+0+0=4$，与右边相符，故 $(2,0)$ 在曲线上。(M1)

Substitute $(x,y)=(2,0)$ into part (a): $\dfrac{dy}{dx}=-\dfrac{2(2)+2(0)}{2(2)+3(0)^{2}}=-\dfrac{4}{4}=-1$. (A1)将 $(x,y)=(2,0)$ 代入 (a) 的结果：$\dfrac{dy}{dx}=-\dfrac{2(2)+2(0)}{2(2)+3(0)^{2}}=-\dfrac{4}{4}=-1$。(A1)

(c) Point-slope form at $(2,0)$(c) 在 $(2,0)$ 处用点斜式 A1

With slope $-1$ through $(2,0)$: $y-0=-1(x-2)$, i.e. $y=-x+2$. (A1)斜率为 $-1$，过点 $(2,0)$：$y-0=-1(x-2)$，即 $y=-x+2$。(A1)

Insight.要点。 Always verify a given point satisfies the curve equation before trusting any downstream slope computation, plugging a point that is not actually on the curve into an implicit-differentiation formula produces a number with no geometric meaning. Notice also that the constant on the right side of the curve equation drops out entirely in part (a): differentiating a constant always gives $0$, so $\dfrac{dy}{dx}$ never depends on it.在信任任何后续斜率计算之前，务必先验证给定点确实满足曲线方程，将不在曲线上的点代入隐函数微分公式所得的数值不具有任何几何意义。另请注意，曲线方程右边的常数在 (a) 部分完全消失：常数求导恒为 $0$，故 $\dfrac{dy}{dx}$ 从不依赖于它。

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FRQ 3MEDIUM 3.3 / 3.4 Inverse Functions3.3 / 3.4 反函数No Calculator[6 marks]

Let $f(x)=x^{3}+2x-1$.设 $f(x)=x^{3}+2x-1$。

Answers:答案：  (a) $f'(x)=3x^{2}+2>0$ always恒成立  ·  (b) $f(1)=2,\ (f^{-1})'(2)=\dfrac{1}{5}$  ·  (c) $g'(1)=1$

(a) Show $f'$ never vanishes or changes sign(a) 说明 $f'$ 恒不为零且不变号 M1·R1

$f'(x)=3x^{2}+2$. Since $x^{2}\ge 0$ for all real $x$, $3x^{2}\ge 0$, so $f'(x)=3x^{2}+2\ge 2>0$ for every real $x$. (M1)$f'(x)=3x^{2}+2$。因对所有实数 $x$ 均有 $x^{2}\ge 0$，故 $3x^{2}\ge 0$，即 $f'(x)=3x^{2}+2\ge 2>0$ 对所有实数 $x$ 恒成立。(M1)

Because $f'(x)>0$ everywhere, $f$ is strictly increasing on all of $\mathbb{R}$, and a strictly increasing function can never repeat an output value, so $f$ is one-to-one. (R1)因 $f'(x)>0$ 处处成立，故 $f$ 在整个 $\mathbb{R}$ 上严格递增，而严格递增函数不可能重复取同一函数值，故 $f$ 为一一映射。(R1)

(b) Evaluate $f(1)$, then apply the inverse-derivative formula(b) 求 $f(1)$，再套用反函数导数公式 M1·A1

$f(1)=1^{3}+2(1)-1=1+2-1=2$, so $f^{-1}(2)=1$. (M1)$f(1)=1^{3}+2(1)-1=1+2-1=2$，故 $f^{-1}(2)=1$。(M1)

From (a), $f'(1)=3(1)^{2}+2=5$, so $(f^{-1})'(2)=\dfrac{1}{f'(f^{-1}(2))}=\dfrac{1}{f'(1)}=\dfrac{1}{5}$. (A1)由 (a)，$f'(1)=3(1)^{2}+2=5$，故 $(f^{-1})'(2)=\dfrac{1}{f'(f^{-1}(2))}=\dfrac{1}{f'(1)}=\dfrac{1}{5}$。(A1)

(c) Differentiate the composition $\arctan(f(x))$(c) 对复合函数 $\arctan(f(x))$ 求导 M1·A1

By the chain rule, $g'(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}$. (M1)由链式法则，$g'(x)=\dfrac{f'(x)}{1+[f(x)]^{2}}$。(M1)

Substitute $x=1$, using $f(1)=2$ and $f'(1)=5$ from (b): $g'(1)=\dfrac{5}{1+2^{2}}=\dfrac{5}{5}=1$. (A1)代入 $x=1$，利用 (b) 中 $f(1)=2$ 与 $f'(1)=5$：$g'(1)=\dfrac{5}{1+2^{2}}=\dfrac{5}{5}=1$。(A1)

Insight.要点。 All three parts reuse the same two numbers, $f(1)=2$ and $f'(1)=5$, first to certify invertibility, then to evaluate the inverse derivative, then to evaluate a further composition: compute the building blocks once and carry them forward rather than recomputing. This "$f'>0$ everywhere $\Rightarrow$ one-to-one" argument is the standard justification whenever a cubic (or any function) is claimed invertible without solving for $f^{-1}$ explicitly.三个部分反复使用同两个数值，$f(1)=2$ 与 $f'(1)=5$：先用于证明可逆性，再用于求反函数导数，最后用于求另一复合函数的导数。应一次性算出这些基本量，之后直接沿用，而非重复计算。"$f'>0$ 处处成立 $\Rightarrow$ 一一映射"是断言一个三次函数（或任意函数）可逆而无需显式求出 $f^{-1}$ 时的标准论证方式。

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FRQ 4HARD 4.4 Related Rates (Cone)4.4 相关变化率（圆锥体）Preview of Unit 4Calculator[8 marks]

Water is poured into a right-circular-cone tank (vertex down), radius $4$ ft at the top, height $6$ ft, at $2$ ft³/min.以 $2$ ft³/min 的速率向顶点朝下的正圆锥形水箱注水，该水箱顶口半径为 $4$ 英尺，高度为 $6$ 英尺。

Answers:答案：  (a) $V=\dfrac{4\pi}{27}h^{3}$  ·  (b) $\dfrac{dh}{dt}=\dfrac{1}{2\pi}\approx 0.159$ ft/minft/min  ·  (c) faster at $h=2$ ft$h=2$ 英尺时更快

(a) Use similar triangles to eliminate $r$(a) 用相似三角形消去 $r$ M1·M1·A1

The water's cross-section is always similar to the full tank's cross-section, so its radius $r$ and height $h$ stay in the fixed ratio of the tank itself: $\dfrac{r}{h}=\dfrac{4}{6}=\dfrac{2}{3}$, so $r=\dfrac{2}{3}h$. (M1)水面的横截面始终与水箱整体的横截面相似，故其半径 $r$ 与高度 $h$ 保持水箱自身的固定比例：$\dfrac{r}{h}=\dfrac{4}{6}=\dfrac{2}{3}$，故 $r=\dfrac{2}{3}h$。(M1)

Substitute into the cone volume formula $V=\dfrac{1}{3}\pi r^{2}h$: (M1)代入圆锥体积公式 $V=\dfrac{1}{3}\pi r^{2}h$：(M1)

$$ V=\frac{1}{3}\pi\left(\frac{2}{3}h\right)^{2}h=\frac{1}{3}\pi\cdot\frac{4}{9}h^{2}\cdot h=\frac{4\pi}{27}h^{3}. $$

(A1)

(b) Differentiate $V(h)$ with respect to $t$, then solve for $\dfrac{dh}{dt}$(b) 对 $V(h)$ 关于 $t$ 求导，再解出 $\dfrac{dh}{dt}$ M1·M1·A1

Differentiate part (a) with respect to $t$: $\dfrac{dV}{dt}=\dfrac{4\pi}{27}\cdot 3h^{2}\dfrac{dh}{dt}=\dfrac{4\pi}{9}h^{2}\dfrac{dh}{dt}$. (M1)对 (a) 关于 $t$ 求导：$\dfrac{dV}{dt}=\dfrac{4\pi}{27}\cdot 3h^{2}\dfrac{dh}{dt}=\dfrac{4\pi}{9}h^{2}\dfrac{dh}{dt}$。(M1)

At $h=3$, $h^{2}=9$, so $\dfrac{dV}{dt}=\dfrac{4\pi}{9}(9)\dfrac{dh}{dt}=4\pi\dfrac{dh}{dt}$. Substituting $\dfrac{dV}{dt}=2$: $2=4\pi\dfrac{dh}{dt}$. (M1)当 $h=3$ 时，$h^{2}=9$，故 $\dfrac{dV}{dt}=\dfrac{4\pi}{9}(9)\dfrac{dh}{dt}=4\pi\dfrac{dh}{dt}$。代入 $\dfrac{dV}{dt}=2$：$2=4\pi\dfrac{dh}{dt}$。(M1)

So $\dfrac{dh}{dt}=\dfrac{2}{4\pi}=\dfrac{1}{2\pi}\approx 0.159$ ft/min. (A1)故 $\dfrac{dh}{dt}=\dfrac{2}{4\pi}=\dfrac{1}{2\pi}\approx 0.159$ ft/min。(A1)

(c) Compare $\dfrac{dh}{dt}$ at the two heights using the general relationship(c) 用一般关系式比较两高度处的 $\dfrac{dh}{dt}$ M1·R1

From part (b)'s relationship, $\dfrac{dh}{dt}=\dfrac{dV/dt}{(4\pi/9)h^{2}}=\dfrac{9\,(dV/dt)}{4\pi h^{2}}$, so with $dV/dt$ fixed at $2$ ft³/min, $\dfrac{dh}{dt}$ is inversely proportional to $h^{2}$: at $h=2$, $\dfrac{dh}{dt}=\dfrac{4.5}{\pi(4)}\approx 0.358$ ft/min; at $h=4$, $\dfrac{dh}{dt}=\dfrac{4.5}{\pi(16)}\approx 0.089$ ft/min. (M1)由 (b) 部分的关系式，$\dfrac{dh}{dt}=\dfrac{dV/dt}{(4\pi/9)h^{2}}=\dfrac{9\,(dV/dt)}{4\pi h^{2}}$，故当 $dV/dt$ 固定为 $2$ ft³/min 时，$\dfrac{dh}{dt}$ 与 $h^{2}$ 成反比：当 $h=2$ 时，$\dfrac{dh}{dt}=\dfrac{4.5}{\pi(4)}\approx 0.358$ ft/min；当 $h=4$ 时，$\dfrac{dh}{dt}=\dfrac{4.5}{\pi(16)}\approx 0.089$ ft/min。(M1)

Since $0.358>0.089$, the water height is rising faster when $h=2$ ft: the tank is narrower there, so the same inflow rate raises the (smaller) cross-section faster. (R1)因 $0.358>0.089$，故当 $h=2$ 英尺时水位上升更快：此处水箱更窄，同样的进水速率会使（较小的）横截面上升得更快。(R1)

Insight.要点。 Substituting similar triangles to write $V$ as a function of the single variable $h$ before differentiating is the standard move for every cone-shaped related-rates problem, it avoids ever needing $\dfrac{dr}{dt}$ at all. Part (c) generalizes the pattern from Q16: whenever $\dfrac{dh}{dt}\propto 1/h^{2}$ (or any inverse power of $h$), the rate is always largest at the smallest $h$, no separate computation is needed once that proportionality is established.在求导前先用相似三角形将 $V$ 化为单变量 $h$ 的函数，是每一道圆锥体相关变化率问题的标准做法，这样便完全无需用到 $\dfrac{dr}{dt}$。(c) 部分推广了 Q16 中的模式：只要 $\dfrac{dh}{dt}\propto 1/h^{2}$（或 $h$ 的任意负幂次），变化率必在 $h$ 最小处取最大值，一旦确立此比例关系，无需再单独计算。

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FRQ 5HARD 3.1 / 3.2 / 3.5 Second Derivative3.1 / 3.2 / 3.5 二阶导数No Calculator[8 marks]

Consider the curve $y^{2}=x^{3}+2x$.考虑曲线 $y^{2}=x^{3}+2x$。

Answers:答案：  (a) $\dfrac{dy}{dx}=\dfrac{3x^{2}+2}{2y}$  ·  (b) $\dfrac{5\sqrt{3}}{6}$  ·  (c) $\dfrac{11\sqrt{3}}{36}$  ·  (d) concave up上凸

(a) Differentiate implicitly, isolating $\dfrac{dy}{dx}$(a) 隐函数求导，解出 $\dfrac{dy}{dx}$ M1·A1

Differentiate both sides: $2y\dfrac{dy}{dx}=3x^{2}+2$. (M1)两边求导：$2y\dfrac{dy}{dx}=3x^{2}+2$。(M1)

So $\dfrac{dy}{dx}=\dfrac{3x^{2}+2}{2y}$. (A1)故 $\dfrac{dy}{dx}=\dfrac{3x^{2}+2}{2y}$。(A1)

(b) Substitute the point into part (a)(b) 将该点代入 (a) 的结果 A1

Confirm $(1,\sqrt3)$ is on the curve: $y^2=3$ and $x^3+2x=1+2=3$, matching. At $(1,\sqrt{3})$: $\dfrac{dy}{dx}=\dfrac{3(1)^{2}+2}{2\sqrt{3}}=\dfrac{5}{2\sqrt{3}}=\dfrac{5\sqrt{3}}{6}$. (A1)验证 $(1,\sqrt3)$ 在曲线上：$y^2=3$，$x^3+2x=1+2=3$，二者相符。在 $(1,\sqrt{3})$ 处：$\dfrac{dy}{dx}=\dfrac{3(1)^{2}+2}{2\sqrt{3}}=\dfrac{5}{2\sqrt{3}}=\dfrac{5\sqrt{3}}{6}$。(A1)

(c) Differentiate part (a)'s equation again(c) 对 (a) 的方程再求导一次 M1·M1·A1

Differentiate $2y\dfrac{dy}{dx}=3x^{2}+2$ with respect to $x$, using the product rule on the left: $2\left(\dfrac{dy}{dx}\right)^{2}+2y\dfrac{d^{2}y}{dx^{2}}=6x$. (M1)对 $2y\dfrac{dy}{dx}=3x^{2}+2$ 关于 $x$ 求导，左边用乘积法则：$2\left(\dfrac{dy}{dx}\right)^{2}+2y\dfrac{d^{2}y}{dx^{2}}=6x$。(M1)

Solve for $\dfrac{d^{2}y}{dx^{2}}$: $\dfrac{d^{2}y}{dx^{2}}=\dfrac{6x-2(dy/dx)^{2}}{2y}=\dfrac{3x-(dy/dx)^{2}}{y}$. At $(1,\sqrt{3})$, using $\dfrac{dy}{dx}=\dfrac{5\sqrt{3}}{6}$ from (b): $\left(\dfrac{dy}{dx}\right)^{2}=\dfrac{25\cdot 3}{36}=\dfrac{25}{12}$. (M1)解出 $\dfrac{d^{2}y}{dx^{2}}$：$\dfrac{d^{2}y}{dx^{2}}=\dfrac{6x-2(dy/dx)^{2}}{2y}=\dfrac{3x-(dy/dx)^{2}}{y}$。在 $(1,\sqrt{3})$ 处，利用 (b) 中 $\dfrac{dy}{dx}=\dfrac{5\sqrt{3}}{6}$：$\left(\dfrac{dy}{dx}\right)^{2}=\dfrac{25\cdot 3}{36}=\dfrac{25}{12}$。(M1)

So $\dfrac{d^{2}y}{dx^{2}}=\dfrac{3(1)-25/12}{\sqrt{3}}=\dfrac{11/12}{\sqrt{3}}=\dfrac{11}{12\sqrt{3}}=\dfrac{11\sqrt{3}}{36}$. (A1)故 $\dfrac{d^{2}y}{dx^{2}}=\dfrac{3(1)-25/12}{\sqrt{3}}=\dfrac{11/12}{\sqrt{3}}=\dfrac{11}{12\sqrt{3}}=\dfrac{11\sqrt{3}}{36}$。(A1)

(d) Read the sign of $\dfrac{d^{2}y}{dx^{2}}$(d) 判断 $\dfrac{d^{2}y}{dx^{2}}$ 的符号 A1·R1

From part (c), $\dfrac{d^{2}y}{dx^{2}}=\dfrac{11\sqrt{3}}{36}>0$. (A1)由 (c)，$\dfrac{d^{2}y}{dx^{2}}=\dfrac{11\sqrt{3}}{36}>0$。(A1)

Since the second derivative is positive at $(1,\sqrt{3})$, the curve is concave up there. (R1)因二阶导数在 $(1,\sqrt{3})$ 处为正，故曲线在该点处上凸。(R1)

Insight.要点。 This question is Q14's pattern with a cubic instead of a circle: differentiate implicitly once for $\dfrac{dy}{dx}$, plug that expression's numeric value into a second implicit differentiation for $\dfrac{d^{2}y}{dx^{2}}$, never resubstitute the original curve equation by mistake. Carrying the exact fraction $\dfrac{5\sqrt3}{6}$ forward from (b) into (c), rather than rounding early, is what keeps the final answer exact.本题是将 Q14 的模式套用于三次曲线而非圆：先隐函数求导一次得 $\dfrac{dy}{dx}$，再将该表达式的具体数值代入第二次隐函数求导以得到 $\dfrac{d^{2}y}{dx^{2}}$，切勿误将原曲线方程再次代入。将 (b) 中的精确分数 $\dfrac{5\sqrt3}{6}$ 原样带入 (c)，而非提前取近似值，正是保持最终答案精确的关键。

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