Sections 1 to 7: $e^x$ and $\ln x$, trig derivatives, general $a^x$ and $\log_a x$, logarithmic differentiation, inverse-trig derivatives, hyperbolic functions, synthesis第 1 至 7 节：$e^x$ 与 $\ln x$、三角导数、一般 $a^x$ 与 $\log_a x$、对数求导法、反三角导数、双曲函数、综合CALC I
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PART I · CORE TECHNIQUES核心技巧Computational fluency · 28 marks计算熟练度 · 28分
Differentiating Transcendental Functions超越函数的求导
Differentiate each function fully. Simplify your answer to a single unsplit fraction or a standard factored form. State which rule (chain, product, quotient) you apply at each step that combines two or more rules.对各函数完整求导。将答案化简为单一未拆分分式或标准因式分解形式。说明在每个结合两条或更多规则的步骤中所应用的规则（链式、乘积、商法则）。
Differentiate each function with respect to $x$. You may use the standard formulas directly.对各函数关于 $x$ 求导。可直接使用标准公式。
(a) $f(x)=3e^{x}-5\ln x+2$ [2]
(b) $g(x)=4^{x}+\log_{4}x$ [2]
(c) $h(x)=x^{2}e^{x}-\ln(x^{3})$ [2]
Q2MEDIUMCOREderivatives of $\sin$, $\cos$, $\tan$ and the other trig functions$\sin$、$\cos$、$\tan$ 及其他三角函数的导数[6 marks]
Differentiate each function. For (c), derive the result by writing $\cot x = \cos x/\sin x$ and applying the quotient rule; do not quote $-\csc^{2}x$ without justification.对各函数求导。对于 (c)，将 $\cot x = \cos x/\sin x$ 代入并应用商法则推导结果，不得直接引用 $-\csc^{2}x$ 而不加说明。
(a) $f(x)=3\sin x-2\cos x+\tan x$ [2]
(b) $g(x)=x\cos x+\sec x$ [2]
(c) $h(x)=\cot x$ (derive via quotient rule)（用商法则推导）[2]
Q3MEDIUMCOREchain rule with exponential and logarithmic functions链式法则与指数及对数函数[8 marks]
Differentiate each function. Identify the outer and inner functions before computing.对各函数求导。计算前先确定外层函数和内层函数。
(a) $f(x)=e^{3x^{2}-1}$ [2]
(b) $g(x)=\ln(\sin x)$, for $\sin x>0$其中 $\sin x>0$[2]
(c) $p(x)=5^{\cos x}$ [2]
(d) $q(x)=\ln\!\left(\dfrac{e^{x}+1}{e^{x}-1}\right)$, for $e^{x}>1$其中 $e^{x}>1$[2]
Q4HARDCOREproduct and quotient rules combined with trig and exponential乘积法则与商法则结合三角及指数函数[8 marks]
Differentiate each function fully. Simplify where possible; do not leave unexpanded products in the numerator of a quotient-rule result.对各函数完整求导。尽可能化简；商法则结果的分子中不得保留未展开的乘积。
(a) $f(x)=e^{x}\sin x \cos x$ [3]
(b) $g(x)=\dfrac{x^{2}e^{x}}{\ln x}$, for $x>0,\,x\ne 1$其中 $x>0,\,x\ne 1$[3]
(c) $h(x)=e^{-x^{2}}\tan x$ [2]
PART II · DEFINITIONS AND PROOF定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Deriving the Transcendental Derivative Formulas推导超越函数导数公式
Every proof must include: the definition or identity used to start; all intermediate algebraic steps; a clearly stated conclusion. Marks are awarded for the logic of the argument, not only the final answer. Avoid circular reasoning: do not use a formula in its own derivation.每道证明题必须包含：所用的定义或恒等式；所有中间代数步骤；明确陈述的结论。评分依据论证逻辑，而非仅看最终答案。避免循环论证：不得在推导公式时使用该公式本身。
Q5HARDPROOFderiving $\frac{d}{dx}\ln x = \frac{1}{x}$ and $\frac{d}{dx}a^{x}=a^{x}\ln a$推导 $\frac{d}{dx}\ln x = \frac{1}{x}$ 与 $\frac{d}{dx}a^{x}=a^{x}\ln a$[8 marks]
Prove both derivative formulas from first principles. For (a), differentiate the identity $e^{\ln x}=x$ implicitly. For (b), write $a^{x}=e^{x\ln a}$ and apply the chain rule together with the result of (a).从基本原理证明两个导数公式。对于 (a)，对恒等式 $e^{\ln x}=x$ 进行隐函数求导。对于 (b)，将 $a^{x}=e^{x\ln a}$ 代入并结合 (a) 的结果应用链式法则。
(a)Starting from $e^{\ln x}=x$ (valid for $x>0$), differentiate both sides with respect to $x$ and hence prove $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$.从 $e^{\ln x}=x$（对 $x>0$ 成立）出发，对两边关于 $x$ 求导，由此证明 $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$。[4]
(b)Using the representation $a^{x}=e^{x\ln a}$ (where $a>0$, $a\ne 1$), prove that $\dfrac{d}{dx}a^{x}=a^{x}\ln a$. State clearly where the chain rule is applied.利用表达式 $a^{x}=e^{x\ln a}$（其中 $a>0$，$a\ne 1$），证明 $\dfrac{d}{dx}a^{x}=a^{x}\ln a$。明确说明链式法则的应用之处。[4]
Q6HARDPROOFderiving inverse-trig derivatives by implicit differentiation通过隐函数求导推导反三角函数导数[10 marks]
Derive each inverse-trig derivative formula rigorously. In each case, introduce the inverse relation, differentiate implicitly, and use a Pythagorean identity to eliminate the trig function in favour of $x$.严格推导每个反三角函数导数公式。在每种情况下，引入反函数关系，进行隐函数求导，并利用毕达哥拉斯恒等式将三角函数转化为 $x$ 的表达式。
(a)Let $y=\arcsin x$ with $x\in(-1,1)$ and $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. Starting from $\sin y = x$, prove that $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^{2}}}$. (You must state why $\cos y > 0$ on the given domain.)设 $y=\arcsin x$，其中 $x\in(-1,1)$，$y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$。从 $\sin y = x$ 出发，证明 $\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^{2}}}$。（必须说明在给定定义域上 $\cos y > 0$ 的原因。）[5]
(b)Let $y=\arctan x$ with $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. Starting from $\tan y = x$, prove that $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^{2}}$. State the Pythagorean identity you use to simplify $\sec^{2}y$.设 $y=\arctan x$，其中 $y\in\!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$。从 $\tan y = x$ 出发，证明 $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^{2}}$。说明用于化简 $\sec^{2}y$ 的毕达哥拉斯恒等式。[5]
Q7HARDPROOFderiving $\frac{d}{dx}\sin x = \cos x$ from the limit definition从极限定义推导 $\frac{d}{dx}\sin x = \cos x$[8 marks]
Use the limit definition of the derivative. You may assume the two standard limits $\displaystyle\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\displaystyle\lim_{h\to 0}\frac{\cos h-1}{h}=0$ as given facts (you proved $\lim_{h\to 0}\frac{\sin h}{h}=1$ in Unit A1).使用导数的极限定义。可将两个标准极限 $\displaystyle\lim_{h\to 0}\frac{\sin h}{h}=1$ 和 $\displaystyle\lim_{h\to 0}\frac{\cos h-1}{h}=0$ 作为已知事实（你在第A1单元中已证明 $\lim_{h\to 0}\frac{\sin h}{h}=1$）。
(a)Write out $f'(x)$ for $f(x)=\sin x$ using the definition $f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, expand $\sin(x+h)$ using the angle-addition formula, and rearrange to isolate the two standard limits. Hence prove $\dfrac{d}{dx}\sin x = \cos x$.利用定义 $f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ 写出 $f(x)=\sin x$ 的 $f'(x)$，用角和公式展开 $\sin(x+h)$，整理以分离两个标准极限，由此证明 $\dfrac{d}{dx}\sin x = \cos x$。[5]
(b)Using the result of (a) and the identity $\cos x = \sin\!\left(\tfrac{\pi}{2}-x\right)$, deduce that $\dfrac{d}{dx}\cos x = -\sin x$ without repeating the limit argument.利用 (a) 的结果和恒等式 $\cos x = \sin\!\left(\tfrac{\pi}{2}-x\right)$，推导 $\dfrac{d}{dx}\cos x = -\sin x$，无需重复极限论证。[3]
PART III · APPLICATIONS AND SYNTHESIS应用与综合Extended problems · 28 marks综合拓展题 · 28分
Logarithmic Differentiation, Hyperbolic Functions, and Multi-Rule Problems对数求导法、双曲函数与多规则综合题
Set up each problem cleanly. Carry exact forms through intermediate steps. Where logarithmic differentiation is required, take $\ln$ of both sides before differentiating; the final answer must be expressed in terms of the original function $y$, not $\ln y$.清晰建立每道题的解题框架。中间步骤保持精确形式。需要对数求导时，先对两边取 $\ln$ 再求导；最终答案必须用原函数 $y$ 表示，而非 $\ln y$。
Q8HARDAPPLIEDlogarithmic differentiation: $x^x$ and a multi-factor product对数求导法：$x^x$ 与多因子乘积[8 marks]
Differentiate each function using logarithmic differentiation. For (a), the power and the base are both variable; the standard power rule does not apply.用对数求导法对各函数求导。对于 (a)，幂次和底数均为变量，标准幂法则不适用。
(a)Find $\dfrac{dy}{dx}$ where $y = x^{x}$, for $x>0$.求 $\dfrac{dy}{dx}$，其中 $y = x^{x}$，$x>0$。[4]
(b)Find $\dfrac{dy}{dx}$ where $y = \dfrac{x^{3}\sqrt{x+1}}{e^{2x}(1+x^{2})}$, for $x>-1$. Use logarithmic differentiation to avoid the full quotient rule; simplify to a single fraction multiplied by $y$.求 $\dfrac{dy}{dx}$，其中 $y = \dfrac{x^{3}\sqrt{x+1}}{e^{2x}(1+x^{2})}$，$x>-1$。使用对数求导法以避免完整商法则；化简为单一分式乘以 $y$。[4]
Q9HARDAPPLIEDhyperbolic functions and identities; optimisation with $e^x$双曲函数与恒等式；$e^x$ 的最优化[10 marks]
The hyperbolic functions are defined by $\sinh x = \dfrac{e^{x}-e^{-x}}{2}$ and $\cosh x = \dfrac{e^{x}+e^{-x}}{2}$. Treat them as functions of $e^{x}$ and $e^{-x}$; do not quote derivative formulas without deriving them in this question.双曲函数定义为 $\sinh x = \dfrac{e^{x}-e^{-x}}{2}$，$\cosh x = \dfrac{e^{x}+e^{-x}}{2}$。将其视为 $e^{x}$ 和 $e^{-x}$ 的函数；本题中不得直接引用导数公式而不加推导。
(a)Prove that $\dfrac{d}{dx}\sinh x = \cosh x$ and $\dfrac{d}{dx}\cosh x = \sinh x$ directly from the definitions.直接从定义证明 $\dfrac{d}{dx}\sinh x = \cosh x$ 且 $\dfrac{d}{dx}\cosh x = \sinh x$。[3]
(c)Find all critical points of $g(x) = xe^{-x}$ and determine whether each is a local maximum, a local minimum, or neither. Justify using the second derivative.求 $g(x) = xe^{-x}$ 的所有极值点，判断各点是极大值、极小值还是两者均非。用二阶导数判别法说明理由。[4]
This question combines inverse-trig derivatives with the chain and product rules, and applies them geometrically.本题将反三角导数与链式法则、乘积法则结合，并进行几何应用。
(a)Differentiate $f(x)=\arctan\!\left(\dfrac{2x}{1-x^{2}}\right)$ and simplify your answer. (Hint: first compute $f'$ via the chain rule; after simplification the answer is a constant multiple of a simple rational function.)对 $f(x)=\arctan\!\left(\dfrac{2x}{1-x^{2}}\right)$ 求导并化简答案。（提示：先用链式法则计算 $f'$；化简后答案为某简单有理函数的常数倍。）[4]
(b)Differentiate $p(x) = x\arcsin x + \sqrt{1-x^{2}}$ and simplify fully.对 $p(x) = x\arcsin x + \sqrt{1-x^{2}}$ 求导并完全化简。[3]
(c)Let $r(x) = e^{\arctan x}$. Find the equation of the tangent line to the curve $y = r(x)$ at the point where $x = 1$. Give exact coordinates.设 $r(x) = e^{\arctan x}$。求曲线 $y = r(x)$ 在 $x = 1$ 处的切线方程，给出精确坐标。[3]