Sections 1 to 7: the limit definition, the derivative as a function, slope and rate interpretations, differentiability vs continuity, first-principles computation, higher-order derivatives, where differentiability fails第 1 至 7 节：极限定义，导函数，斜率与变化率解释，可微性与连续性，第一性原理计算，高阶导数，可微性失效之处CALC I
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PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算流畅度 · 28分
Limit Definition and the Derivative as a Function极限定义与导函数
Show all working. Use the limit definition $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ or $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ exactly as specified. Simplify the difference quotient algebraically before taking the limit.展示所有计算过程。严格按照题目要求使用极限定义 $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ 或 $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$。在取极限前先对差商进行代数化简。
Q1MEDIUMCORElimit definition at a point, both forms极限定义求一点处导数，两种形式[6 marks]
Use the limit definition of the derivative to compute each of the following.用导数的极限定义计算以下各项。
Q2MEDIUMCOREderivative as a function, piecewise and one-sided derivatives导函数，分段函数与单侧导数[6 marks]
Let $g(x) = |2x - 4|$.设 $g(x) = |2x - 4|$。
(a)Write $g$ explicitly as a piecewise-linear function.将 $g$ 明确写成分段函数的形式。[1]
(b)Using the limit definition, compute $g'_{-}(2)$ (the left-hand derivative at $x=2$) and $g'_{+}(2)$ (the right-hand derivative at $x=2$).使用极限定义，计算 $g'_{-}(2)$（$x=2$ 处的左侧导数）和 $g'_{+}(2)$（$x=2$ 处的右侧导数）。[4]
(c)State, with a one-sentence justification, whether $g$ is differentiable at $x=2$.用一句话说明理由，判断 $g$ 在 $x=2$ 处是否可微。[1]
Q3HARDCOREdifferentiability of a piecewise function, parameter matching分段函数的可微性，参数匹配[8 marks]
Define定义
$$ h(x) = \begin{cases} ax^{2}+b, & x \le 1, \\ 3x-2, & x > 1. \end{cases} $$
(a)State the two conditions on $a$ and $b$ that are necessary for $h$ to be differentiable at $x=1$. (One condition comes from continuity; one comes from the derivatives matching.)给出 $a$ 和 $b$ 须满足的两个条件，使得 $h$ 在 $x=1$ 处可微。（一个条件来自连续性，另一个来自导数匹配。）[2]
(b)Compute $h'_{-}(1)$ (using the left formula) and $h'_{+}(1)$ (using the right formula) in terms of $a$, then set them equal to find $a$.用含 $a$ 的表达式计算 $h'_{-}(1)$（使用左侧公式）和 $h'_{+}(1)$（使用右侧公式），然后令二者相等求出 $a$。[3]
(c)Use the continuity condition to find $b$, and write down the complete function $h$.利用连续性条件求 $b$，并写出完整的函数 $h$。[3]
Q4HARDCOREhigher-order derivatives, velocity and acceleration高阶导数，速度与加速度[8 marks]
A particle moves along a straight line. Its position at time $t \ge 0$ (in seconds) is given by
$$ s(t) = t^{3} - 6t^{2} + 9t + 2, $$
where $s$ is measured in metres. Use the limit definition to find $s'(t)$ from first principles at an arbitrary $t$, and then use differentiation rules (which you may quote) for higher-order derivatives.一质点沿直线运动。其在时刻 $t \ge 0$（单位：秒）时的位置为
$$ s(t) = t^{3} - 6t^{2} + 9t + 2, $$
其中 $s$ 以米为单位。用极限定义从第一性原理求任意 $t$ 处的 $s'(t)$，然后使用微分法则（可直接引用）求高阶导数。
(a)Compute $s'(t)$ using $s'(t)=\displaystyle\lim_{h\to 0}\frac{s(t+h)-s(t)}{h}$. Expand fully and collect terms before taking the limit.使用 $s'(t)=\displaystyle\lim_{h\to 0}\frac{s(t+h)-s(t)}{h}$ 计算 $s'(t)$。取极限前须完整展开并合并同类项。[4]
(b)Find the acceleration function $a(t)=s''(t)$ by differentiating $s'(t)$.对 $s'(t)$ 求导，得到加速度函数 $a(t)=s''(t)$。[2]
(c)Find all values of $t$ at which the particle is momentarily at rest, and determine whether the particle is accelerating or decelerating at each such time.求质点瞬间静止时所有 $t$ 的值，并判断在每个该时刻质点是在加速还是减速。[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
First Principles, Differentiability, and Continuity第一性原理，可微性与连续性
These items are graded on the logic of the argument, not just the final line. In a proof, every implication must be justified. When computing a derivative from the limit definition, you must simplify the difference quotient algebraically before invoking the limit; writing "take $h\to 0$" on an unsimplified quotient earns no credit.这些题目按论证逻辑评分，而不仅看最终结论。在证明中，每一步推导都必须有依据。用极限定义计算导数时，必须在取极限前对差商进行代数化简；对未化简的差商直接写"令 $h\to 0$"不得分。
Let $f(x)=\dfrac{1}{x^{2}}$, defined for $x\ne 0$.设 $f(x)=\dfrac{1}{x^{2}}$，定义域为 $x\ne 0$。
(a)Using $f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, derive a formula for $f'(x)$. Show every algebraic step: combine the fractions over a common denominator, cancel $h$, then take the limit.使用 $f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$，推导 $f'(x)$ 的公式。展示每一步代数运算：通分，约去 $h$，再取极限。[5]
(b)Verify your formula by checking it against the power rule $\dfrac{d}{dx}[x^{n}]=nx^{n-1}$.用幂法则 $\dfrac{d}{dx}[x^{n}]=nx^{n-1}$ 验证你的公式。[1]
(c)Find the equation of the tangent line to $y=\dfrac{1}{x^{2}}$ at $x=-1$, and the equation of the normal line at the same point.求曲线 $y=\dfrac{1}{x^{2}}$ 在 $x=-1$ 处的切线方程，以及同一点处的法线方程。[2]
This item requires a rigorous proof using limits. Do not appeal to intuition or graphs.本题要求用极限给出严格证明。不得诉诸直觉或图形。
(a)Prove the theorem: if $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$. Your proof must start from the definition $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ and show that $\lim_{h\to 0}f(a+h)=f(a)$ by writing $f(a+h)-f(a)$ as a product involving the difference quotient and $h$.证明定理：若 $f$ 在 $x=a$ 处可微，则 $f$ 在 $x=a$ 处连续。你的证明必须从定义 $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ 出发，通过将 $f(a+h)-f(a)$ 写成差商与 $h$ 的乘积，证明 $\lim_{h\to 0}f(a+h)=f(a)$。[5]
(b)Explain why the converse is false: construct a specific function that is continuous at a point but not differentiable there, and verify both claims (continuity and non-differentiability) using limits.解释为什么逆命题不成立：构造一个在某点连续但不可微的具体函数，并用极限验证连续性和不可微性这两个结论。[3]
(a)Use the limit definition $f'(a)=\displaystyle\lim_{h\to 0}\frac{(a+h)^{3}-a^{3}}{h}$ to prove that $f'(a)=3a^{2}$ for all $a\in\mathbb{R}$. Expand $(a+h)^{3}$ fully, collect all terms containing $h$ as a factor, cancel $h$, then take the limit.使用极限定义 $f'(a)=\displaystyle\lim_{h\to 0}\frac{(a+h)^{3}-a^{3}}{h}$，证明对所有 $a\in\mathbb{R}$ 有 $f'(a)=3a^{2}$。完整展开 $(a+h)^{3}$，合并所有含 $h$ 的项，约去 $h$，再取极限。[5]
(b)The slope of the tangent to $y=x^{3}$ at a point $(a, a^{3})$ equals the slope of the secant line joining $(a, a^{3})$ to $(a+h, (a+h)^{3})$. Show algebraically that this secant slope equals $3a^{2}+3ah+h^{2}$, and explain what happens to this expression as $h\to 0$.曲线 $y=x^{3}$ 在点 $(a, a^{3})$ 处的切线斜率等于连接 $(a, a^{3})$ 与 $(a+h, (a+h)^{3})$ 的割线斜率。用代数方法证明该割线斜率等于 $3a^{2}+3ah+h^{2}$，并解释当 $h\to 0$ 时该表达式的变化趋势。[3]
(c)At what point(s) on the curve $y=x^{3}$ is the tangent line horizontal? Justify your answer using $f'(a)=3a^{2}$.曲线 $y=x^{3}$ 上哪些点的切线是水平的？用 $f'(a)=3a^{2}$ 说明理由。[2]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Tangent Lines, Rates of Change, and Where Differentiability Fails切线，变化率与可微性失效之处
Set up each problem from the definition. Carry exact values and simplify only at the end. When asked to classify a failure of differentiability, state the one-sided limits of the difference quotient explicitly.从定义出发建立每道题。保持精确值，仅在最后化简。当要求对可微性失效进行分类时，须明确写出差商的单侧极限。
Q8HARDAPPLIEDtangent and normal lines, slope as instantaneous rate切线与法线，斜率作为瞬时变化率[10 marks]
Consider the curve $y = \sqrt{2x+1}$ for $x \ge -\tfrac{1}{2}$.考虑曲线 $y = \sqrt{2x+1}$，其中 $x \ge -\tfrac{1}{2}$。
(a)Using the limit definition with the $h\to 0$ form, show that $\dfrac{d}{dx}\!\left[\sqrt{2x+1}\right] = \dfrac{1}{\sqrt{2x+1}}$. Rationalise the numerator of the difference quotient to clear the indeterminate form.使用 $h\to 0$ 形式的极限定义，证明 $\dfrac{d}{dx}\!\left[\sqrt{2x+1}\right] = \dfrac{1}{\sqrt{2x+1}}$。对差商的分子有理化，以消去不定式。[4]
(b)Find the equation of the tangent line to the curve at the point where $x=4$.求曲线在 $x=4$ 处的切线方程。[3]
(c)Find the equation of the normal line to the curve at the same point $x=4$, and show that the tangent and normal lines are perpendicular.求曲线在同一点 $x=4$ 处的法线方程，并证明切线与法线互相垂直。[3]
Q9HARDAPPLIEDinstantaneous rate of change, average rate, velocity interpretation瞬时变化率，平均变化率，速度解释[10 marks]
The volume of a spherical balloon is $V(r)=\tfrac{4}{3}\pi r^{3}$, where $r$ is the radius in centimetres.球形气球的体积为 $V(r)=\tfrac{4}{3}\pi r^{3}$，其中 $r$ 为半径，单位为厘米。
(a)Compute the average rate of change of $V$ with respect to $r$ over the interval $[2, 2+h]$, simplifying your expression as a polynomial in $h$.计算 $V$ 关于 $r$ 在区间 $[2, 2+h]$ 上的平均变化率，将结果化简为关于 $h$ 的多项式。[3]
(b)Take the limit as $h\to 0$ to find the instantaneous rate of change $V'(r)$ at a general radius $r$. Interpret $V'(r)$ geometrically (your answer should be a familiar formula).令 $h\to 0$ 取极限，求一般半径 $r$ 处的瞬时变化率 $V'(r)$。对 $V'(r)$ 给出几何解释（答案应为一个熟悉的公式）。[3]
(c)At $r=3$ cm, the radius is increasing at $0.5$ cm/s. Use the interpretation of $V'(3)$ as a rate to estimate the instantaneous rate at which the volume is increasing at that moment. Give the units of your answer.在 $r=3$ 厘米时，半径以 $0.5$ 厘米/秒的速度增大。利用 $V'(3)$ 作为变化率的解释，估算该时刻体积增大的瞬时速率。请给出答案的单位。[2]
(d)The average rate of change of $V$ over $[r_{0}, r_{0}+h]$ equals the instantaneous rate at $r_{0}$ only in the limit $h\to 0$. Explain in one sentence what the average rate represents geometrically on the graph of $V$ versus $r$.$V$ 在 $[r_{0}, r_{0}+h]$ 上的平均变化率只有在 $h\to 0$ 的极限下才等于 $r_{0}$ 处的瞬时变化率。用一句话解释平均变化率在 $V$ 关于 $r$ 的图像上的几何意义。[2]
For each function below, determine whether it is differentiable at the indicated point. If it is not, classify the failure as a corner, a cusp, or a vertical tangent, and support your classification by computing the relevant one-sided limits of the difference quotient.对以下每个函数，判断其在指定点处是否可微。如果不可微，将失效类型分类为角点、尖点或竖直切线，并通过计算差商的相关单侧极限来支持你的分类。
(a) $f(x) = |x^{2}-1|$ at $x=1$.在 $x=1$ 处。[3]
(b) $g(x) = x^{2/3}$ at $x=0$.在 $x=0$ 处。[3]
(c)Sketch the qualitative shape of $f$ near $x=1$ and $g$ near $x=0$ that makes the respective failure of differentiability visually evident. Label the key feature (corner or cusp) on each sketch.画出 $f$ 在 $x=1$ 附近和 $g$ 在 $x=0$ 附近的定性图形，使各自的可微性失效直观可见。在每幅图上标注关键特征（角点或尖点）。[2]