Sections 1 to 7: power/sum/constant-multiple rules, product rule, quotient rule, chain rule, combining rules, polynomials and rational functions, pitfalls第 1 至 7 节：幂法则/求和法则/常数倍法则、乘积法则、商法则、链式法则、综合运用、多项式与有理函数、常见错误CALC I
Name:姓名：Date:日期：
PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算熟练度 · 28分
Applying the Differentiation Rules运用微分法则
Show all working. Identify the rule(s) you are applying at each step. Simplify your answer fully unless the question says otherwise.展示完整解题过程。在每一步中注明所使用的法则。除非题目另有说明，否则将答案化简到最简形式。
Q1MEDIUMCOREpower, constant-multiple, and sum rules幂法则、常数倍法则与求和法则[6 marks]
Differentiate each function. State the rule applied at each step and simplify fully.对每个函数求导。在每一步中注明所用法则，并将结果化简到最简形式。
(b) $g(x)=x^{4}\!\left(2x^{2}-5\right)^{3}$ (treat $\left(2x^{2}-5\right)^{3}$ as a single factor; do not expand the cube)（将 $\left(2x^{2}-5\right)^{3}$ 视为一个整体因子，不得展开立方）[3]
(c) $h(x)=\left(x^{2}+3\right)\!\left(x^{2}-3\right)$ using the product rule, then verify by expanding first and differentiating directly.用乘积法则求导，再通过先展开后直接求导的方式验证结果。[2]
Q3HARDCOREchain rule, including nested chains链式法则，含嵌套链式[8 marks]
Differentiate each composite function. Identify the outer and inner functions explicitly before applying the chain rule.对每个复合函数求导。在运用链式法则之前，明确指出外层函数与内层函数。
(a) $f(x)=\left(3x^{4}-2x+1\right)^{6}$ [2]
(b) $g(x)=\sqrt{5x^{3}+\sqrt{x}}$ (a chain within a chain: identify all layers)（链中套链：请逐层指出所有层次）[3]
(c) $h(x)=\left(\dfrac{x^{2}+1}{x-1}\right)^{4}$ (chain rule with an inner quotient; do not expand)（内层含商式的链式法则，不得展开）[3]
Q4HARDCOREquotient rule商法则[6 marks]
Differentiate using the quotient rule. Simplify numerators fully but leave denominators in factored form.用商法则求导。将分子化简到最简形式，分母保持因式分解的形式。
(c) $h(x)=\dfrac{x^{3}}{x^{2}+3}$, and find all values of $x$ at which $h'(x)=0$.并求所有使 $h'(x)=0$ 成立的 $x$ 值。[2]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Deriving the Rules from First Principles从第一性原理推导法则
These items are graded on the logic of the argument, not just the final line. Every proof must start from the limit definition of the derivative and proceed by named algebraic steps. Quoting a rule you are asked to prove earns zero marks.这些题目按论证逻辑评分，而非仅看最终结论。每项证明必须从导数的极限定义出发，并逐步注明代数步骤的名称。直接引用所求证法则者不得分。
Q5HARDPROOFproduct rule from the limit definition由极限定义推导乘积法则[10 marks]
Let $f$ and $g$ be differentiable at $x$. The goal is to prove that $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$.设 $f$ 与 $g$ 在 $x$ 处可微。目标是证明 $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$。
(a)Write down the difference quotient $\dfrac{(fg)(x+h)-(fg)(x)}{h}$ and then add and subtract the quantity $f(x+h)g(x)$ in the numerator. Factor the result into two separate difference quotients.写出差商 $\dfrac{(fg)(x+h)-(fg)(x)}{h}$，然后在分子中加减量 $f(x+h)g(x)$，并将结果整理为两个独立差商之和。[4]
(b)Take the limit as $h\to 0$. Identify which standard limits give $f'(x)$ and $g'(x)$, and explain why $f(x+h)\to f(x)$ is needed. Write the conclusion as $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$.令 $h\to 0$ 取极限。指出哪些标准极限给出 $f'(x)$ 和 $g'(x)$，并解释为何需要 $f(x+h)\to f(x)$。写出结论 $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$。[3]
(c)Use the product rule you just proved to differentiate $p(x)=\left(x^{3}+1\right)\!\left(2x^{2}-3x\right)$ without expanding, and simplify.用刚刚证明的乘积法则对 $p(x)=\left(x^{3}+1\right)\!\left(2x^{2}-3x\right)$ 求导，不得展开，并化简结果。[3]
Q6HARDPROOFpower rule for integer n by mathematical induction用数学归纳法证明整数次幂法则[8 marks]
The power rule states: for any positive integer $n$, $\dfrac{d}{dx}\!\left(x^{n}\right)=nx^{n-1}$.幂法则：对任意正整数 $n$，有 $\dfrac{d}{dx}\!\left(x^{n}\right)=nx^{n-1}$。
(a)Prove the base case $n=1$ directly from the limit definition $f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$.直接由极限定义 $f'(x)=\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ 证明基础情形 $n=1$。[3]
(b)Assume the rule holds for $n=k$, i.e. $\dfrac{d}{dx}\!\left(x^{k}\right)=kx^{k-1}$. Write $x^{k+1}=x\cdot x^{k}$, apply the product rule (proved in Q5), and the inductive hypothesis to obtain $\dfrac{d}{dx}\!\left(x^{k+1}\right)=(k+1)x^{k}$.假设法则对 $n=k$ 成立，即 $\dfrac{d}{dx}\!\left(x^{k}\right)=kx^{k-1}$。将 $x^{k+1}=x\cdot x^{k}$，利用（Q5 中已证的）乘积法则及归纳假设，推导出 $\dfrac{d}{dx}\!\left(x^{k+1}\right)=(k+1)x^{k}$。[3]
(c)State the conclusion of the induction, and then use the power rule to find the second derivative $f''(x)$ and third derivative $f'''(x)$ of $f(x)=x^{5}-3x^{3}+2x$.写出归纳的结论，然后用幂法则求 $f(x)=x^{5}-3x^{3}+2x$ 的二阶导数 $f''(x)$ 与三阶导数 $f'''(x)$。[2]
Q7MEDIUMPROOFquotient rule derived from the product rule由乘积法则推导商法则[8 marks]
Let $f$ and $g$ be differentiable at $x$ with $g(x)\ne 0$. Denote $Q(x)=\dfrac{f(x)}{g(x)}$.设 $f$ 与 $g$ 在 $x$ 处可微且 $g(x)\ne 0$。记 $Q(x)=\dfrac{f(x)}{g(x)}$。
(a)Write $f(x)=Q(x)\cdot g(x)$ and differentiate both sides using the product rule. Rearrange to isolate $Q'(x)$. You may use the fact that if $g$ is differentiable and nonzero then $\dfrac{d}{dx}\!\left(\dfrac{1}{g(x)}\right)=-\dfrac{g'(x)}{[g(x)]^{2}}$.写出 $f(x)=Q(x)\cdot g(x)$，对两边用乘积法则求导，再整理以分离出 $Q'(x)$。可以使用以下事实：若 $g$ 可微且不为零，则 $\dfrac{d}{dx}\!\left(\dfrac{1}{g(x)}\right)=-\dfrac{g'(x)}{[g(x)]^{2}}$。[3]
(b)Write the result in the standard form $Q'(x)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$ and identify which sign in the numerator distinguishes the quotient rule from the product rule.将结果写成标准形式 $Q'(x)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$，并指出分子中哪个符号将商法则与乘积法则区分开来。[3]
(c)Use the quotient rule to differentiate $r(x)=\dfrac{x^{4}-1}{x^{2}+1}$ and factor the numerator of $r'(x)$ completely.用商法则对 $r(x)=\dfrac{x^{4}-1}{x^{2}+1}$ 求导，并将 $r'(x)$ 的分子完全因式分解。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合应用题 · 28分
Combining Rules, Tangent Lines, and Analysis综合运用法则、切线与分析
Set up each problem cleanly. Where a derivative is used to find a geometric quantity, state the connection explicitly. Carry exact values through intermediate steps.清晰地建立每道题的解题框架。若用导数求几何量，需明确说明二者的联系。中间步骤保留精确值。
Q8HARDAPPLIEDchain rule combined with product rule链式法则与乘积法则的综合运用[8 marks]
Let $F(x)=x^{3}\!\left(x^{2}+4\right)^{5}$.设 $F(x)=x^{3}\!\left(x^{2}+4\right)^{5}$。
(a)Identify the two factors and state which rule(s) are needed. Then differentiate $F(x)$ using the product rule together with the chain rule on the second factor. Do not expand $\left(x^{2}+4\right)^{5}$.指出两个因子，并说明需要哪些法则。然后对第二个因子运用链式法则，结合乘积法则对 $F(x)$ 求导。不得展开 $\left(x^{2}+4\right)^{5}$。[2]
(b)Factor $F'(x)$ completely by extracting the common factor $x^{2}\!\left(x^{2}+4\right)^{4}$. Write the result in the form $F'(x)=x^{2}\!\left(x^{2}+4\right)^{4}\!\left(Ax^{2}+B\right)$ and identify $A$ and $B$.提取公因子 $x^{2}\!\left(x^{2}+4\right)^{4}$，对 $F'(x)$ 完全因式分解。将结果写成 $F'(x)=x^{2}\!\left(x^{2}+4\right)^{4}\!\left(Ax^{2}+B\right)$ 的形式，并确定 $A$ 与 $B$。[3]
(c)Determine all values of $x$ at which $F'(x)=0$ and, for each, classify whether the corresponding point on the graph of $F$ is a local maximum, local minimum, or neither (use sign analysis of $F'$, not a second-derivative test).求所有使 $F'(x)=0$ 成立的 $x$ 值，并对每个值判断 $F$ 图像上对应点是极大值点、极小值点还是两者均非（用 $F'$ 的符号分析，不得用二阶导数判别法）。[3]
Q9HARDAPPLIEDhorizontal tangents of a rational function有理函数的水平切线[10 marks]
(a)State the domain of $f$ and explain why the denominator is never zero.写出 $f$ 的定义域，并解释分母为何永不为零。[2]
(b)Use the quotient rule to find $f'(x)$ and simplify it to the form $\dfrac{16x}{(x^{2}+4)^{2}}$. Show the unsimplified numerator before cancelling.用商法则求 $f'(x)$，将其化简为 $\dfrac{16x}{(x^{2}+4)^{2}}$ 的形式。在约分之前，展示未化简的分子。[3]
(c)Find all points on the graph of $f$ where the tangent line is horizontal. Give coordinates.求 $f$ 图像上所有切线水平的点，给出坐标。[3]
(d)The function $f$ has horizontal asymptotes as $x\to\pm\infty$. Find them and state whether $f$ actually reaches these values or only approaches them.函数 $f$ 在 $x\to\pm\infty$ 时有水平渐近线。求出渐近线，并说明 $f$ 是否真正取到这些值，或仅趋近于它们。[2]
Q10HARDAPPLIEDcombining all rules: tangent line and common pitfalls综合所有法则：切线与常见错误[10 marks]
Let $\displaystyle G(x)=\frac{\left(x^{2}+1\right)^{3}}{x^{3}-8}$, defined for $x\ne 2$.设 $\displaystyle G(x)=\frac{\left(x^{2}+1\right)^{3}}{x^{3}-8}$，定义域为 $x\ne 2$。
(a)Differentiate $G(x)$ using the quotient rule, with the chain rule applied to the numerator. Leave your answer in unsimplified form after cancelling any common factor of $h$.用商法则对 $G(x)$ 求导，分子部分使用链式法则。约去公因子 $h$ 后，将答案保留为未化简形式。[3]
(b)Evaluate $G(3)$ and $G'(3)$. Hence write the equation of the tangent line to $y=G(x)$ at $x=3$.计算 $G(3)$ 与 $G'(3)$，由此写出 $y=G(x)$ 在 $x=3$ 处的切线方程。[3]
(c)A student claims: "Since $\left(x^{2}+1\right)^{3}$ is always positive, $G'(x)$ can never be negative." Identify the error in this reasoning and give a specific value of $x$ for which $G'(x)<0$.某学生声称："由于 $\left(x^{2}+1\right)^{3}$ 恒为正，$G'(x)$ 不可能为负。"指出该推理中的错误，并给出一个使 $G'(x)<0$ 的具体 $x$ 值。[2]
(d)A second student differentiates $G$ incorrectly as $G'(x)=\dfrac{3(x^{2}+1)^{2}}{3x^{2}}=\dfrac{(x^{2}+1)^{2}}{x^{2}}$. State the two rules that were misapplied and correct both errors.另一学生将 $G$ 错误地求导为 $G'(x)=\dfrac{3(x^{2}+1)^{2}}{3x^{2}}=\dfrac{(x^{2}+1)^{2}}{x^{2}}$。指出被误用的两条法则，并分别纠正错误。[2]