PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 21 marks不可使用计算器 · 简答题 · 21 分
Section A · Short ResponseA 节 · 简答题
Show every differentiation step. Name the rule you use (chain, product, quotient, log) at the start of each computation. Simplify the final expression and factor common terms where possible. No calculator permitted.写出每一步求导过程。每次计算开头注明所用法则（链式、乘积、商、对数）。化简最终表达式，能提公因式就提。不可使用计算器。
Let $g(x) = \ln(\cos x)$ on the interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.设 $g(x) = \ln(\cos x)$，定义于 $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$。
(a)Explain why $g$ is well-defined on this interval.说明 $g$ 在该区间上为何有定义。[1]
(b)Apply the chain rule to show that $g'(x) = -\tan x$.用链式法则证明 $g'(x) = -\tan x$。[3]
(c)Find $g''(x)$ and evaluate it at $x = 0$.求 $g''(x)$，并在 $x = 0$ 处求值。[2]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分
Section B · Extended ResponseB 节 · 长答题
Show every implicit-differentiation step. When you differentiate a term containing $y$, write the $\dfrac{dy}{dx}$ factor on the same line so the marker sees the chain rule being applied. Collect $\dfrac{dy}{dx}$ terms on one side before solving.写出每一步隐函数求导过程。求导含 $y$ 的项时，将 $\dfrac{dy}{dx}$ 因子写在同一行，让评卷人看到链式法则的应用。整理时把含 $\dfrac{dy}{dx}$ 的项归到一边再求解。
A curve $C$ in the $xy$-plane is defined by the equation$xy$ 平面上的曲线 $C$ 由方程
$$ x^{3} + 3xy + y^{3} \;=\; 1. $$
(a)Verify that the point $P(0,\, 1)$ lies on $C$.验证点 $P(0,\, 1)$ 在 $C$ 上。[1]
(b)Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.两边对 $x$ 求导，把 $y$ 视为 $x$ 的函数。[3]
(c)Hence show that $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$, stating the values of $(x, y)$ for which this expression is undefined.由此证明 $\dfrac{dy}{dx} = -\dfrac{x^{2} + y}{x + y^{2}}$，并指出该表达式无意义的 $(x, y)$ 值。[3]
(d)Find the equation of the tangent line to $C$ at $P(0,\, 1)$, giving the answer in the form $y = mx + c$.求曲线 $C$ 在 $P(0,\, 1)$ 处的切线方程，写成 $y = mx + c$。[2]
(e)State the equation of the tangent line in the form $ax + by = c$ with integer coefficients.将该切线写成整数系数形式 $ax + by = c$。[2]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 16 marks可使用计算器 · 混合题型 · 16 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. Use it to confirm derivatives (numerical derivative nDeriv) and to evaluate slopes. Always cross-check by writing the closed-form derivative by hand before plugging in numerical values. Give exact answers where reasonable; otherwise round to $3$ s.f.需要图形计算器（GDC）。可用其确认导数（数值导数 nDeriv）并计算切线斜率。代入数值前务必先用手算写出导数闭式以交叉验证。能精确求解的题给精确答案，否则保留 $3$ 位有效数字。
Let $f(x) = x \, e^{-x^{2}/2}$ for $x \in \mathbb{R}$.设 $f(x) = x \, e^{-x^{2}/2}$，$x \in \mathbb{R}$。
(a)Use the product rule and the chain rule to show that $f'(x) = (1 - x^{2})\, e^{-x^{2}/2}$.用乘积法则与链式法则证明 $f'(x) = (1 - x^{2})\, e^{-x^{2}/2}$。[3]
(b)Find the exact $x$-coordinates of the stationary points of $f$, and confirm them by graphing $f$ and $f'$ on your GDC.求 $f$ 的驻点的精确 $x$ 坐标，并用 GDC 绘制 $f$ 与 $f'$ 进行确认。[2]
(c)Use your GDC to evaluate the slope $f'(0.5)$ to $3$ s.f., and verify the value by substituting into the closed-form derivative from (a).用 GDC 求斜率 $f'(0.5)$（保留 $3$ 位有效数字），并代入 (a) 中的闭式导数加以验证。[2]
Q7HARDPaper 2AHL 5.13 Tangent to an Implicit Curve (HL)[9 marks]
Consider the curve $K$ defined by $x^{2} + xy + y^{2} = 3$.考虑由方程 $x^{2} + xy + y^{2} = 3$ 定义的曲线 $K$。
(a)Verify that the point $Q(1,\, 1)$ lies on $K$.验证点 $Q(1,\, 1)$ 在 $K$ 上。[1]
(b)Differentiate the defining equation implicitly with respect to $x$.将定义方程对 $x$ 隐式求导。[3]
(c)Hence find $\left.\dfrac{dy}{dx}\right|_{(1,1)}$ and the equation of the tangent line to $K$ at $Q$.由此求 $\left.\dfrac{dy}{dx}\right|_{(1,1)}$ 以及曲线 $K$ 在 $Q$ 处的切线方程。[3]
(d)Use your GDC to plot $K$ (implicit plot or by solving for $y$ as the two branches $y = \tfrac{1}{2}\bigl(-x \pm \sqrt{12 - 3x^{2}}\bigr)$) and confirm that the tangent at $Q$ has the slope found in (c). State briefly what you observed.用 GDC 绘制 $K$（隐式绘图，或解出两支 $y = \tfrac{1}{2}\bigl(-x \pm \sqrt{12 - 3x^{2}}\bigr)$），确认 $Q$ 处的切线斜率与 (c) 一致。简述所观察到的结果。[2]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分
Paper 3 · HL Extended Problem第三卷 · HL 长题探究
A graphing calculator is required. Method marks dominate. For logarithmic differentiation, you must write the line $\ln y = (\text{expression in } x)$ before differentiating, and you must reinstate $y$ after solving for $\dfrac{dy}{dx}$.需要图形计算器（GDC）。方法分占主导。对数求导法须在求导之前写出 $\ln y = (\text{含 } x \text{ 的表达式})$；解出 $\dfrac{dy}{dx}$ 后须把 $y$ 代回。
In this problem you investigate functions of the form $y = u(x)^{v(x)}$ where the base and exponent both depend on $x$. The rule $\dfrac{d}{dx}x^{n} = n x^{n-1}$ (constant exponent) and $\dfrac{d}{dx}a^{x} = a^{x}\ln a$ (constant base) do not apply directly; logarithmic differentiation is the standard tool.本题考查形如 $y = u(x)^{v(x)}$ 的函数，其中底数与指数均依赖于 $x$。法则 $\dfrac{d}{dx}x^{n} = n x^{n-1}$（指数为常数）与 $\dfrac{d}{dx}a^{x} = a^{x}\ln a$（底数为常数）不能直接套用；对数求导法是标准工具。
(a)Let $y = x^{\sin x}$ for $x > 0$. By taking $\ln$ of both sides and differentiating implicitly, show that设 $y = x^{\sin x}$，$x > 0$。两边取 $\ln$ 并隐式求导，证明
$$ \frac{dy}{dx} \;=\; x^{\sin x}\left(\cos x \, \ln x + \frac{\sin x}{x}\right). $$
[4]
(b)Use your GDC to evaluate $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{2}$ to $3$ s.f., and verify the value by substituting into the formula in (a).用 GDC 求 $\dfrac{dy}{dx}$ 在 $x = \dfrac{\pi}{2}$ 处的值（$3$ 位有效数字），并代入 (a) 中的公式验证。[2]
(c)Now let $z = (x^{2} + 1)^{x^{2}}$ for $x \in \mathbb{R}$. Apply logarithmic differentiation to find $\dfrac{dz}{dx}$ in a fully simplified form.现设 $z = (x^{2} + 1)^{x^{2}}$，$x \in \mathbb{R}$。用对数求导法求 $\dfrac{dz}{dx}$ 的完全化简形式。[4]
(d)More generally, let $w = u(x)^{v(x)}$ with $u(x) > 0$ and both $u, v$ differentiable. Show that更一般地，设 $w = u(x)^{v(x)}$，$u(x) > 0$，$u, v$ 均可微。证明
$$ \frac{dw}{dx} \;=\; u(x)^{v(x)}\left[v'(x)\, \ln u(x) + \frac{v(x)\, u'(x)}{u(x)}\right]. $$
Hence verify that the answers to (a) and (c) are consistent with this general formula.由此验证 (a) 与 (c) 的答案与该一般公式一致。[5]