Unit 3: Composite, Implicit
& Inverse Functions
单元 3：复合函数、隐函数
与反函数
Master the chain rule and its extensions to implicit, inverse, and higher-order derivatives — essential groundwork for integration techniques in Unit 6 and beyond.
掌握链式法则及其在隐函数、反函数与高阶导数上的延伸——这是单元 6 及之后积分技巧的必备基础。
The Chain Rule
链式法则（chain rule）
The chain rule is the single most important differentiation technique you will learn in this unit — and arguably in the entire course. It allows you to differentiate composite functions: functions built by nesting one function inside another.
链式法则是本单元——甚至可以说整门课程中最重要的一种求导（differentiate）技巧。它让你能够对复合函数（composite function）求导：复合函数是把一个函数嵌套在另一个函数里构造而成的。
If $y = f(g(x))$, then the derivative is found by differentiating the outer function evaluated at the inner function, then multiplying by the derivative of the inner function:
若 $y = f(g(x))$，则其导数（derivative）通过先对外函数（outer function）在内函数处求导，再乘以内函数（inner function）的导数来得到：
In Leibniz notation, if $y = f(u)$ and $u = g(x)$, then:
在莱布尼茨记法（Leibniz notation）下，若 $y = f(u)$ 且 $u = g(x)$，则：
A practical mnemonic: every time you apply the chain rule, say to yourself "times the derivative of what's inside." This simple habit prevents the most common chain rule error — forgetting to multiply by the inner derivative.
一个实用的口诀：每次使用链式法则时，对自己念一句"再乘以里面那一层的导数。"这个小习惯能够防止链式法则中最常见的错误——忘记乘以内函数的导数。
Think of the chain rule as an extension of unit analysis. Just as $\dfrac{\text{psi}}{\text{min}} = \dfrac{\text{psi}}{\text{m}} \cdot \dfrac{\text{m}}{\text{min}}$, the Leibniz notation $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ "cancels" the intermediate variable $u$.
可以把链式法则看作单位分析的推广。正如 $\dfrac{\text{psi}}{\text{min}} = \dfrac{\text{psi}}{\text{m}} \cdot \dfrac{\text{m}}{\text{min}}$ 那样，莱布尼茨记法 $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ 把中间变量 $u$ "约掉了"。
Decomposing Composite Functions
分解复合函数
Before applying the chain rule, you must identify which function is the outer function and which is the inner function. Practice decomposing expressions into their component layers:
在使用链式法则之前，必须先识别哪一部分是外函数，哪一部分是内函数。多练习把一个表达式拆解成它的各个嵌套层：
| Function $y$函数 $y$ | Outer $f(u)$外函数 $f(u)$ | Inner $u = g(x)$内函数 $u = g(x)$ | $\frac{dy}{dx}$ |
|---|---|---|---|
| $\sin(4x)$ | $\sin(u)$ | $4x$ | $\cos(4x) \cdot 4$ |
| $(\sin x)^4$ | $u^4$ | $\sin x$ | $4(\sin x)^3 \cdot \cos x$ |
| $e^{x^2}$ | $e^u$ | $x^2$ | $e^{x^2} \cdot 2x$ |
| $\sqrt{3x+1}$ | $\sqrt{u}$ | $3x+1$ | $\dfrac{3}{2\sqrt{3x+1}}$ |
| $\tan(2x-1)$ | $\tan(u)$ | $2x-1$ | $\sec^2(2x-1)\cdot 2$ |
| $\ln(\cos x)$ | $\ln(u)$ | $\cos x$ | $\dfrac{-\sin x}{\cos x} = -\tan x$ |
Nested Chains
嵌套链式法则
Some functions have multiple layers of composition. Apply the chain rule from the outermost layer inward, multiplying each inner derivative as you go.
有些函数包含多层复合（composition）。要从最外层向内逐层使用链式法则，每深入一层就乘以那一层内函数的导数。
Find $\dfrac{d}{dx}\bigl[\sin(e^{3x})\bigr]$.
求 $\dfrac{d}{dx}\bigl[\sin(e^{3x})\bigr]$。
Adjust the parameter $a$ in $y = \sin(ax)$ to see how the chain rule scales the derivative. Fixed axes make it easier to compare how the function and derivative change as $a$ varies.
调整 $y = \sin(ax)$ 中的参数 $a$，观察链式法则如何对导数进行缩放。坐标轴固定，便于比较 $a$ 变化时函数及其导数的变化。
Do not confuse $\sin^2 x$ (which means $(\sin x)^2$) with $\sin(x^2)$. The outer and inner functions are different in each case, producing completely different derivatives.
不要把 $\sin^2 x$（意思是 $(\sin x)^2$）与 $\sin(x^2)$ 弄混。这两种写法的外函数和内函数都不同，得到的导数也完全不同。
On the multiple-choice section, chain-rule distractors almost always include the "forgot the inner derivative" answer. Before you bubble, silently confirm you multiplied by $g'(x)$ — that one factor is the difference between the trap answer and the correct one.
在选择题中，链式法则的干扰项几乎总会包含"忘记内导数"的答案。在填涂之前，先默默确认你乘上了 $g'(x)$——正是这一个因子区分了陷阱答案与正确答案。
Going Deeper — Proving the General Power Rule
拓展——证明一般幂法则
Unit 2's power rule, $\frac{d}{dx}[x^n] = nx^{n-1}$, was justified there for integer $n$. The chain rule (combined with implicit differentiation) lets you extend the proof to any rational exponent $n = p/q$.
第 2 单元的幂法则 $\frac{d}{dx}[x^n] = nx^{n-1}$，当时只对整数 $n$ 给出了论证。借助链式法则（配合隐函数求导），可以把这个证明推广到任意有理指数 $n = p/q$。
Show that $\dfrac{d}{dx}\bigl[x^{p/q}\bigr] = \dfrac{p}{q}x^{(p/q)-1}$ for integers $p, q$ with $q \neq 0$.
证明对整数 $p, q$（$q \neq 0$），$\dfrac{d}{dx}\bigl[x^{p/q}\bigr] = \dfrac{p}{q}x^{(p/q)-1}$。
Step 1 — Set up an implicit equation. Let $y = x^{p/q}$. Raise both sides to the $q$ power to clear the fractional exponent:
第 1 步——建立隐式方程。令 $y = x^{p/q}$。两边取 $q$ 次幂以消去分数指数：
$$y^q = x^p$$Step 2 — Differentiate implicitly. The left side needs the chain rule (power rule on $y$, times $\frac{dy}{dx}$); the right side is an ordinary integer power:
第 2 步——隐式求导。左边需要链式法则（对 $y$ 用幂法则，再乘以 $\frac{dy}{dx}$）；右边是普通的整数次幂：
$$qy^{q-1}\frac{dy}{dx} = px^{p-1}$$Step 3 — Solve for dy/dx.
第 3 步——解出 dy/dx。
$$\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}}$$Step 4 — Substitute $y = x^{p/q}$ back in and simplify the exponent. Since $y^{q-1} = \bigl(x^{p/q}\bigr)^{q-1} = x^{p - p/q}$:
第 4 步——把 $y = x^{p/q}$ 代回并化简指数。因为 $y^{q-1} = \bigl(x^{p/q}\bigr)^{q-1} = x^{p - p/q}$：
$$\frac{dy}{dx} = \frac{p}{q}\cdot\frac{x^{p-1}}{x^{p-p/q}} = \frac{p}{q}x^{(p-1)-(p-p/q)} = \frac{p}{q}x^{(p/q)-1} \qquad\Box$$This is exactly the power rule's formula with $n = p/q$ — the "obvious" pattern you've been using since Unit 2 is not an assumption, it's a theorem, and this is the proof for every rational exponent.
这正是把 $n = p/q$ 代入幂法则得到的公式——你从第 2 单元起就在使用的这条"显然"规律，其实并非假设，而是一条定理，这里就是它对所有有理指数成立的证明。
Implicit Differentiation
隐函数求导（implicit differentiation）
Not every relationship between $x$ and $y$ can be neatly written as $y = f(x)$. When $x$ and $y$ appear together in an equation such as $x^2 + y^2 = 25$, the function is defined implicitly. Implicit differentiation lets you find $\frac{dy}{dx}$ without solving for $y$ first.
并非每一种 $x$ 与 $y$ 的关系都能整齐地写成 $y = f(x)$ 的形式。当 $x$ 与 $y$ 同时出现在像 $x^2 + y^2 = 25$ 这样的方程里时，函数是隐式地被定义的。隐函数求导让你不必先解出 $y$ 就能求 $\frac{dy}{dx}$。
The Procedure
操作步骤
The key insight is that the chain rule applies to $y$ because $y$ depends on $x$. Whenever you differentiate a term containing $y$, you must multiply by $\frac{dy}{dx}$.
关键的认识是：链式法则对 $y$ 适用，因为 $y$ 依赖于 $x$。每当你对含有 $y$ 的项求导时，都必须乘以一个因子 $\frac{dy}{dx}$。
1. Differentiate every term on both sides with respect to $x$.
2. Whenever you differentiate an expression involving $y$, apply the chain rule — attach a factor of $\frac{dy}{dx}$.
3. Collect all terms containing $\frac{dy}{dx}$ on one side.
4. Factor out $\frac{dy}{dx}$ and solve algebraically.
1. 两边的每一项都对 $x$ 求导。
2. 每当对含有 $y$ 的表达式求导时，都要使用链式法则——附上一个因子 $\frac{dy}{dx}$。
3. 把所有含 $\frac{dy}{dx}$ 的项整理到方程的一侧。
4. 提取公因子 $\frac{dy}{dx}$ 并用代数方法解出。
Find $\dfrac{dy}{dx}$ given $x^2 + y^2 = 25$.
已知 $x^2 + y^2 = 25$，求 $\dfrac{dy}{dx}$。
Move a point around the circle and watch the tangent update from the implicit derivative $\dfrac{dy}{dx}=-\dfrac{x}{y}$. The viewing window stays fixed with equal x/y scaling, and the tangent is extended to the graph edges so the geometry is visually accurate.
沿着圆移动一个点，看切线（tangent line）如何根据隐式导数 $\dfrac{dy}{dx}=-\dfrac{x}{y}$ 实时更新。视图窗口保持固定且 $x$、$y$ 等比例缩放，切线延伸至图边缘，几何关系视觉上严格准确。
Find $\dfrac{dy}{dx}$ given $x^2 y + y^3 = 8$.
已知 $x^2 y + y^3 = 8$，求 $\dfrac{dy}{dx}$。
In expressions like $\dfrac{y}{3y - x^2}$, students sometimes forget that differentiating $y$ with respect to $x$ requires the chain rule. Every $y$ term produces a $\frac{dy}{dx}$ factor — no exceptions.
在像 $\dfrac{y}{3y - x^2}$ 这样的表达式中，学生有时会忘记对 $y$ 关于 $x$ 求导需要使用链式法则。每一个 $y$ 项都要产生一个因子 $\frac{dy}{dx}$——没有例外。
When a free-response question asks for a tangent line, it is really testing two skills at once: implicit differentiation to get the slope, then point-slope form to write the line. Show both steps explicitly — the slope value and the final equation each earn credit on the rubric.
当自由作答题要求切线时，它其实同时考查两项技能：用隐函数求导求斜率，再用点斜式写出直线。把两步都清楚写出——斜率的数值和最终方程在评分标准里各得一分。
Going Deeper — Logarithmic Differentiation
拓展——对数求导法
Implicit differentiation has one more powerful use beyond curves like $x^2+y^2=25$: it can differentiate functions that are awkward for the ordinary rules — a variable raised to a variable power ($x^x$), or a long product/quotient of several factors. The trick is to take $\ln$ of both sides first, which turns products into sums, quotients into differences, and powers into multiplication — before you differentiate implicitly.
除了像 $x^2+y^2=25$ 这样的曲线之外，隐函数求导还有一个强大的用途：可以对普通法则难以处理的函数求导——底和指数都含变量的函数（如 $x^x$），或是含多个因子的冗长乘积/商。技巧是先对两边取 $\ln$，这会把乘积变成求和、商变成求差、幂变成乘法——然后再隐式求导。
Two signals: (1) a variable base and a variable exponent in the same expression, or (2) a product/quotient with three or more factors, especially with mixed powers and roots. Neither situation is required knowledge for every AP student, but it's a legitimate technique that turns a multi-rule mess into straightforward term-by-term differentiation, and it does occasionally appear as an enrichment or BC-adjacent problem.
两个信号：(1) 同一表达式中底数和指数都是变量；(2) 含三个或更多因子的乘积/商，尤其是幂和根式混杂的情形。这两种情形并非每位 AP 学生的必修内容，但这是一种合法的技巧，能把一团多法则的混乱变成简单的逐项求导，偶尔也会作为拓展或 BC 相关题目出现。
Preview — Where Implicit Differentiation Goes Next: Related Rates
预告——隐函数求导接下来的应用：相关变化率
Implicit differentiation's most common real-world use isn't a bare equation like $x^2+y^2=25$ — it's a situation where both quantities change with time, and $x$ and $y$ are secretly both functions of a third variable, $t$. Differentiating with respect to $t$ (not $x$) links the two rates of change together.
隐函数求导最常见的实际应用，并不是像 $x^2+y^2=25$ 这样的单独方程——而是两个量都随时间变化的情形，此时 $x$ 与 $y$ 实际上都是第三个变量 $t$ 的函数。对 $t$（而不是 $x$）求导，就能把两个变化率联系起来。
A point moves along the circle $x^2 + y^2 = 25$ so that both $x$ and $y$ change with time $t$. At the instant when $x = 3$, $y = 4$, and $\dfrac{dx}{dt} = 2$, find $\dfrac{dy}{dt}$.
一点沿圆 $x^2 + y^2 = 25$ 运动，$x$ 与 $y$ 都随时间 $t$ 变化。在 $x = 3$、$y = 4$、$\dfrac{dx}{dt} = 2$ 的时刻，求 $\dfrac{dy}{dt}$。
Step 1 — Differentiate with respect to $t$, applying the chain rule to every term with $x$ or $y$.
第 1 步——对 $t$ 求导，对每一个含 $x$ 或 $y$ 的项都使用链式法则。
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$Step 2 — Substitute the known values and solve.
第 2 步——代入已知值并求解。
$$2(3)(2) + 2(4)\frac{dy}{dt} = 0 \;\Longrightarrow\; 12 + 8\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{3}{2}$$This is exactly the same mechanics as finding $\dfrac{dy}{dx}$ implicitly — the only difference is differentiating with respect to $t$ instead of $x$, so every variable (not just $y$) picks up a rate-of-change factor. This idea is developed fully as its own topic later in the course; the point here is that you already have all the differentiation skill required.
这与隐式求 $\dfrac{dy}{dx}$ 的运算过程完全一致——唯一区别是对 $t$ 而不是 $x$ 求导，因此每个变量（不只是 $y$）都会带出一个变化率因子。这个思路会在课程后面作为独立的一个专题完整展开；这里的要点是，你已经具备了所需的全部求导技能。
Differentiating Inverse Functions
反函数求导（inverse function）
If $f$ is a one-to-one differentiable function with inverse $f^{-1}$, you can find the derivative of the inverse without knowing its explicit formula. The relationship rests on the chain rule applied to the identity $f(f^{-1}(x)) = x$.
若 $f$ 是一个一一对应、可导的函数，其反函数为 $f^{-1}$，那么即使不知道反函数的显式公式，也可以求出反函数的导数。这一关系来自对恒等式 $f(f^{-1}(x)) = x$ 使用链式法则。
Every formula in this topic assumes $f$ is one-to-one (each output comes from exactly one input) on the interval in question — otherwise "$f^{-1}$" is not a well-defined function. Graphically, this is the horizontal line test: no horizontal line crosses the graph more than once. A sufficient (and easy to check with calculus) condition is that $f$ is strictly monotonic — always increasing or always decreasing — which happens whenever $f'(x)$ keeps one consistent sign on the interval.
本节的每一个公式都假设 $f$ 在所讨论的区间上一一对应（每个输出恰好来自一个输入）——否则"$f^{-1}$"就不是一个定义良好的函数。从图像上看，这就是水平线检验：任何水平线与图像的交点都不超过一个。一个充分条件（也易于用微积分检验）是 $f$ 严格单调——始终递增或始终递减——只要 $f'(x)$ 在该区间上符号保持一致，就能保证这一点。
derivative of an inverse）Equivalently, if the point $(a, b)$ lies on the graph of $f$ (so $f(a) = b$), then the slope of $f^{-1}$ at $x = b$ is the reciprocal of the slope of $f$ at $x = a$:
等价地，若点 $(a, b)$ 在 $f$ 的图像上（即 $f(a) = b$），那么 $f^{-1}$ 在 $x = b$ 处的斜率（slope）就是 $f$ 在 $x = a$ 处斜率的倒数：
Derivation — Chain Rule on $f(f^{-1}(x)) = x$推导——对 $f(f^{-1}(x)) = x$ 使用链式法则
Suppose $f$ is differentiable and one-to-one on an open interval, with differentiable inverse $f^{-1}$. By definition of inverse, the identity
设 $f$ 在某开区间上可导且一一对应，其反函数 $f^{-1}$ 也可导。由反函数的定义，恒等式
$$f\!\bigl(f^{-1}(x)\bigr) = x$$holds for every $x$ in the range of $f$.
对 $f$ 的值域中的每一个 $x$ 都成立。
Step 1 — Differentiate both sides with respect to $x$. The right side is just $1$. The left side is a composition, so apply the chain rule with outer function $f$ and inner function $f^{-1}(x)$:
第 1 步——两边对 $x$ 求导。右边的导数就是 $1$。左边是一个复合，所以使用链式法则，外函数为 $f$，内函数为 $f^{-1}(x)$：
$$f'\!\bigl(f^{-1}(x)\bigr) \cdot \bigl(f^{-1}\bigr)'(x) \;=\; 1.$$Step 2 — Solve for $(f^{-1})'(x)$. Divide both sides by $f'(f^{-1}(x))$ — this requires $f'(f^{-1}(x)) \ne 0$:
第 2 步——解出 $(f^{-1})'(x)$。两边同时除以 $f'(f^{-1}(x))$——这一步要求 $f'(f^{-1}(x)) \ne 0$：
$$\bigl(f^{-1}\bigr)'(x) \;=\; \frac{1}{f'\!\bigl(f^{-1}(x)\bigr)}.$$Step 3 — Specialize to a point. If $f(a) = b$, then $f^{-1}(b) = a$, so substituting $x = b$ gives the point form:
第 3 步——具体化到一个点。若 $f(a) = b$，则 $f^{-1}(b) = a$，代入 $x = b$ 即得基于点的形式：
$$\bigl(f^{-1}\bigr)'(b) \;=\; \frac{1}{f'(a)}. \qquad\Box$$It is the divide-by-zero step. Geometrically, $f'(a) = 0$ means $f$ has a horizontal tangent at $a$; reflecting that horizontal tangent across $y = x$ produces a vertical tangent on $f^{-1}$, where the derivative does not exist.
它来自避免除以零这一步。从几何上看，$f'(a) = 0$ 意味着 $f$ 在 $a$ 处有水平切线；把这条水平切线沿 $y = x$ 反射，就得到 $f^{-1}$ 上的竖直切线，那里反函数的导数不存在。
The graph of $f^{-1}$ is the reflection of $f$ over the line $y = x$. Reflecting a tangent line swaps the rise and run, which is why the slopes are reciprocals.
$f^{-1}$ 的图像是 $f$ 关于直线 $y = x$ 的反射。反射一条切线会把"升"与"行"互换，因此两者的斜率互为倒数。
Suppose $f(3) = 7$ and $f'(3) = 4$. Find $(f^{-1})'(7)$.
设 $f(3) = 7$ 且 $f'(3) = 4$，求 $(f^{-1})'(7)$。
This formula requires $f'(a) \neq 0$. If the original function has a horizontal tangent at $x = a$, the inverse has a vertical tangent at the corresponding point, and the derivative does not exist there.
这个公式要求 $f'(a) \neq 0$。如果原函数在 $x = a$ 处有水平切线，反函数在对应点就会有竖直切线，那里的导数不存在。
The hardest step is almost never the calculus — it is finding $f^{-1}(x)$, the input that produces the given output. Always solve $f(a) = x$ first, then differentiate. If the problem gives a table, scan the $f(x)$ column for the target value and read off the matching $x$.
最难的一步几乎从来不是微积分本身——而是找到 $f^{-1}(x)$，也就是产生给定输出的那个输入。永远先解 $f(a) = x$，再求导。如果题目给出表格，就在 $f(x)$ 一列里找目标值，读出对应的 $x$。
Going Deeper — Second Derivatives of an Inverse Function
拓展——反函数的二阶导数
The formula $(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}$ is itself a function of $x$ that you can differentiate again — using the chain rule and quotient (or reciprocal) rule together, exactly as in an implicit second-derivative problem.
公式 $(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}$ 本身也是 $x$ 的函数，可以再对它求导——同时用到链式法则和商的法则（或倒数法则），就像隐函数二阶导数问题一样。
Suppose $f(2) = 5$, $f'(2) = 3$, and $f''(2) = 4$. Find $(f^{-1})''(5)$.
设 $f(2) = 5$，$f'(2) = 3$，$f''(2) = 4$，求 $(f^{-1})''(5)$。
Step 1 — Write the reciprocal-rule form of the derivative. Using the reciprocal rule from Topic 2.9 with $g(x) = f'(f^{-1}(x))$:
第 1 步——把导数写成倒数法则的形式。用第 2.9 节的倒数法则，令 $g(x) = f'(f^{-1}(x))$：
$$(f^{-1})''(x) = \frac{d}{dx}\left[\frac{1}{f'(f^{-1}(x))}\right] = -\frac{f''(f^{-1}(x)) \cdot (f^{-1})'(x)}{\bigl[f'(f^{-1}(x))\bigr]^2}$$Step 2 — Evaluate the pieces at $x = 5$. Since $f(2) = 5$, $f^{-1}(5) = 2$. So $f'(f^{-1}(5)) = f'(2) = 3$ and $f''(f^{-1}(5)) = f''(2) = 4$. From Topic 3.3's basic formula, $(f^{-1})'(5) = \dfrac{1}{f'(2)} = \dfrac{1}{3}$.
第 2 步——在 $x = 5$ 处求出各部分的值。由 $f(2) = 5$ 知 $f^{-1}(5) = 2$。故 $f'(f^{-1}(5)) = f'(2) = 3$，$f''(f^{-1}(5)) = f''(2) = 4$。由 3.3 节的基本公式，$(f^{-1})'(5) = \dfrac{1}{f'(2)} = \dfrac{1}{3}$。
Step 3 — Substitute everything in.
第 3 步——把所有结果代入。
$$(f^{-1})''(5) = -\frac{4 \cdot \tfrac{1}{3}}{3^2} = -\frac{4/3}{9} = -\frac{4}{27}$$This kind of problem rewards recognizing the structure — it's the reciprocal/quotient rule and chain rule you already know, applied to a formula you already derived, not a new rule to memorize.
这类题目考察的是识别结构的能力——用到的仍是你已经掌握的倒数/商法则和链式法则，套用在你已经推导过的公式上，而不是要背一条新法则。
Differentiating Inverse Trigonometric Functions
反三角函数（inverse trigonometric function）求导
The derivatives of the inverse trigonometric functions can be derived using implicit differentiation (e.g., let $y = \arcsin(x)$, so $\sin(y) = x$, and differentiate implicitly). You should memorize these results:
反三角函数的导数可以借助隐函数求导得到（例如，令 $y = \arcsin(x)$，于是 $\sin(y) = x$，再用隐函数求导）。下列结果需要背熟：
| Function函数 | Derivative导数 | Domain定义域 |
|---|---|---|
| $\arcsin(x)$ | $\dfrac{1}{\sqrt{1 - x^2}}$ | $-1 < x < 1$ |
| $\arccos(x)$ | $\dfrac{-1}{\sqrt{1 - x^2}}$ | $-1 < x < 1$ |
| $\arctan(x)$ | $\dfrac{1}{1 + x^2}$ | all real $x$全体实数 $x$ |
| $\text{arccot}(x)$ | $\dfrac{-1}{1 + x^2}$ | all real $x$全体实数 $x$ |
| $\text{arcsec}(x)$ | $\dfrac{1}{|x|\sqrt{x^2 - 1}}$ | $|x| > 1$ |
| $\text{arccsc}(x)$ | $\dfrac{-1}{|x|\sqrt{x^2 - 1}}$ | $|x| > 1$ |
Notice that the "co-" functions ($\arccos$, $\text{arccot}$, $\text{arccsc}$) each have derivatives that are the negatives of their non-"co" counterparts. Memorize three formulas and add a minus sign for the cofunctions.
注意到带 "co-" 前缀的反三角函数（$\arccos$、$\text{arccot}$、$\text{arccsc}$）的导数都是其对应不带 "co-" 函数导数的相反数。只需记三条公式，再给余函数加上负号即可。
Each cofunction identity ($\arcsin x + \arccos x = \tfrac{\pi}{2}$, $\arctan x + \text{arccot}\,x = \tfrac{\pi}{2}$, $\text{arcsec}\,x + \text{arccsc}\,x = \tfrac{\pi}{2}$) differentiates to "derivative of the plain function $+$ derivative of the co-function $= 0$," since $\tfrac{\pi}{2}$ is a constant. That forces the co-function's derivative to be the exact negative of its partner's — the same argument used above for $\arccos$ generalizes to all three pairs at once, so you only ever need to derive (or memorize) half the table.
每一对余函数恒等式（$\arcsin x + \arccos x = \tfrac{\pi}{2}$、$\arctan x + \text{arccot}\,x = \tfrac{\pi}{2}$、$\text{arcsec}\,x + \text{arccsc}\,x = \tfrac{\pi}{2}$）求导后都变为"普通函数的导数 + 余函数的导数 = 0"，因为 $\tfrac{\pi}{2}$ 是常数。这就迫使余函数的导数恰好是其配对函数导数的相反数——前面对 $\arccos$ 用过的这个论证，可以一次性推广到全部三对函数，所以你实际上只需要推导（或记住）表格的一半。
Derivation — $\dfrac{d}{dx}[\arcsin x]$ via implicit differentiation推导——用隐函数求导求 $\dfrac{d}{dx}[\arcsin x]$
Step 1 — Set up. Let $y = \arcsin(x)$. By definition of the principal branch, $\sin y = x$ with $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$.
第 1 步——设定。令 $y = \arcsin(x)$。按主值支（principal branch）的定义，$\sin y = x$，且 $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$。
Step 2 — Differentiate implicitly with respect to $x$. Chain rule on the left:
第 2 步——两边对 $x$ 隐式求导。左侧使用链式法则：
$$\begin{aligned} \cos y \cdot \frac{dy}{dx} &= 1 \\[4pt] \frac{dy}{dx} &= \frac{1}{\cos y}. \end{aligned}$$Step 3 — Convert $\cos y$ back to $x$. By the Pythagorean identity, $\cos^2 y = 1 - \sin^2 y = 1 - x^2$, so $\cos y = \pm\sqrt{1 - x^2}$.
第 3 步——把 $\cos y$ 换回 $x$。由毕达哥拉斯恒等式，$\cos^2 y = 1 - \sin^2 y = 1 - x^2$，故 $\cos y = \pm\sqrt{1 - x^2}$。
Step 4 — Pick the correct branch. On $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$, cosine is non-negative. Take the $+$ sign: $\cos y = \sqrt{1 - x^2}$.
第 4 步——选取正确的符号支。在 $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$ 上，余弦非负，取正号：$\cos y = \sqrt{1 - x^2}$。
Step 5 — Conclude.
第 5 步——得出结论。
$$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1.$$At $x = \pm 1$, $\cos y = 0$, and the formula divides by zero. Geometrically, $\arcsin$ has a vertical tangent at $\pm 1$ — its derivative is undefined there.
当 $x = \pm 1$ 时，$\cos y = 0$，公式会除以零。从几何上看，$\arcsin$ 在 $\pm 1$ 处有竖直切线——那里的导数没有定义。
Derivation — $\dfrac{d}{dx}[\arccos x]$ via implicit differentiation推导——用隐函数求导求 $\dfrac{d}{dx}[\arccos x]$
Same recipe. Let $y = \arccos(x)$, so $\cos y = x$ with $0 \le y \le \pi$.
套路相同。令 $y = \arccos(x)$，于是 $\cos y = x$，且 $0 \le y \le \pi$。
Differentiate implicitly:
隐式求导：
$$\begin{aligned} -\sin y \cdot \frac{dy}{dx} &= 1 \\[4pt] \frac{dy}{dx} &= -\frac{1}{\sin y}. \end{aligned}$$Convert $\sin y$: $\sin^2 y = 1 - \cos^2 y = 1 - x^2$, so $\sin y = \pm\sqrt{1 - x^2}$.
把 $\sin y$ 换回 $x$：$\sin^2 y = 1 - \cos^2 y = 1 - x^2$，故 $\sin y = \pm\sqrt{1 - x^2}$。
Pick the branch: on $0 \le y \le \pi$, sine is non-negative, so $\sin y = +\sqrt{1 - x^2}$.
选取符号支：在 $0 \le y \le \pi$ 上，正弦非负，所以 $\sin y = +\sqrt{1 - x^2}$。
Conclude:
得出结论：
$$\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1.$$$\arcsin x + \arccos x = \tfrac{\pi}{2}$ for all $x \in [-1, 1]$. Differentiating both sides gives $\dfrac{d}{dx}[\arcsin x] + \dfrac{d}{dx}[\arccos x] = 0$, so $\arccos$'s derivative is the negative of $\arcsin$'s — consistent with our result.
对所有 $x \in [-1, 1]$ 都有 $\arcsin x + \arccos x = \tfrac{\pi}{2}$。两边求导得 $\dfrac{d}{dx}[\arcsin x] + \dfrac{d}{dx}[\arccos x] = 0$，所以 $\arccos$ 的导数就是 $\arcsin$ 导数的相反数——与我们得到的结果一致。
With the Chain Rule
与链式法则结合
On the AP Exam, these functions almost always appear with a non-trivial inner function. Apply the chain rule:
在 AP 考试中，这些函数几乎总是带有一个非平凡的内函数出现。要使用链式法则：
arctan）与链式法则Find $\dfrac{d}{dx}\bigl[\arcsin(3x^2)\bigr]$.
求 $\dfrac{d}{dx}\bigl[\arcsin(3x^2)\bigr]$。
Vary $a$ to see how the chain rule factor changes the steepness of $\arctan(ax)$ and the height of its derivative peak. The graph window stays fixed for direct visual comparison.
改变 $a$，观察链式法则的因子如何改变 $\arctan(ax)$ 的陡峭程度以及其导数峰值的高度。图像窗口固定，便于直接比较。
Inverse trig derivative questions are graded almost entirely on whether you picked the right formula from the table and correctly applied the chain rule to the inner function — there's rarely any algebraic simplification expected afterward. Spend your time double-checking the sign (arccos, arccot, arccsc are negative) and the inner derivative, not searching for a "nicer" final form.
反三角函数求导题的得分几乎完全取决于你是否从表中选对了公式，并对内函数正确使用了链式法则——之后通常不需要再做代数化简。把时间花在核对符号（arccos、arccot、arccsc 均为负）和内导数上，而不是去找一个"更好看"的最终形式。
Selecting Procedures for Calculating Derivatives
选择计算导数的方法
By this point in the course, you have accumulated a large toolkit of differentiation rules: the power rule, product rule, quotient rule, chain rule, implicit differentiation, and inverse function derivatives. The challenge on the AP Exam is often not executing a rule, but recognizing which rules to apply and in what order.
学到这里，你已经积累了一整套求导工具：幂法则（power rule）、乘积法则（product rule）、商的法则（quotient rule）、链式法则、隐函数求导以及反函数的导数。AP 考试的挑战往往不在于使用某一条法则，而在于识别要用哪些法则、以什么顺序使用。
Decision Framework
决策框架
When you encounter a derivative problem, ask yourself the following questions in order:
遇到求导题时，按以下顺序自问以下几个问题：
Simplify first. Expand, factor, rewrite radicals as powers, or simplify fractions before differentiating.
先化简。在求导之前先展开、因式分解、把根号写成幂，或化简分式。
Product? If the expression is a product of two or more functions, use the product rule — often together with the chain rule.
是乘积吗？如果表达式是两个或多个函数的乘积，就使用乘积法则——通常还要配合链式法则。
Quotient? Use the quotient rule, or rewrite the expression as a product with a negative exponent when that makes the derivative cleaner.
是商吗？使用商的法则；当负指数的乘积形式让导数更整齐时，也可以把表达式改写成那种形式。
Function inside a function? That is a composition, so apply the chain rule.
函数套在函数里？这就是一个复合，使用链式法则。
$x$ and $y$ mixed together? Use implicit differentiation.
$x$ 与 $y$ 混在一起？使用隐函数求导。
Derivative of an inverse? Use the inverse function derivative formula.
反函数的导数？使用反函数的导数公式。
Many AP problems do not fit into just one rule. Read the expression in layers, then combine the correct tools in the correct order.
很多 AP 题目无法只用一条法则解决。要分层阅读表达式，然后按正确顺序组合使用合适的工具。
Outside structure: a product. Inside, a composition. Use the product rule, and inside one term use the chain rule.
外层结构：乘积。内层包含一个复合。先使用乘积法则，再对其中某一项使用链式法则。
Overall structure: a quotient. Inside the numerator, a composition. Use the quotient rule together with the chain rule.
整体结构：商。分子中含一个复合。结合使用商的法则与链式法则。
Because the two variables are mixed together, start with implicit differentiation. Then differentiate any product term with the product rule.
由于两个变量混在一起，先用隐函数求导。然后对任何乘积项使用乘积法则。
Recognize the outer function as an inverse trig function, then apply its derivative formula and multiply by the derivative of the inside using the chain rule.
识别出外函数是反三角函数，使用相应的导数公式，再借助链式法则乘以内函数的导数。
Find $\dfrac{dy}{dx}$ for $y = x^2 \sqrt{\sin(3x)}$.
求 $y = x^2 \sqrt{\sin(3x)}$ 的 $\dfrac{dy}{dx}$。
Step 1 — Identify the outermost structure. This is a product of $x^2$ and $\sqrt{\sin(3x)}$, so the product rule governs the top level.
第 1 步——识别最外层结构。这是 $x^2$ 与 $\sqrt{\sin(3x)}$ 的乘积，所以最外层要用乘积法则。
Step 2 — Look inside the second factor. $\sqrt{\sin(3x)} = [\sin(3x)]^{1/2}$ is a composition three layers deep: an outer power, then sine, then $3x$. That calls for the chain rule applied twice.
第 2 步——查看第二个因子的内部。$\sqrt{\sin(3x)} = [\sin(3x)]^{1/2}$ 是一个三层的复合：外层是幂，再是正弦，再是 $3x$。这需要两次使用链式法则。
$$\frac{d}{dx}\left[\sin(3x)\right]^{1/2} = \tfrac{1}{2}\left[\sin(3x)\right]^{-1/2} \cdot \cos(3x) \cdot 3 = \frac{3\cos(3x)}{2\sqrt{\sin(3x)}}$$Step 3 — Assemble with the product rule.
第 3 步——套用乘积法则组合。
$$\frac{dy}{dx} = 2x\sqrt{\sin(3x)} + x^2 \cdot \frac{3\cos(3x)}{2\sqrt{\sin(3x)}}$$Notice that no single rule from the toolkit solved this alone — the skill being tested is the decomposition, not any one differentiation formula.
注意，工具箱里没有任何单一法则能独自解决这道题——真正考察的能力是分解结构，而不是某一条求导公式本身。
Calculating Higher-Order Derivatives
计算高阶导数（higher-order derivative）
A higher-order derivative is simply the derivative of a derivative. Differentiating $f'(x)$ gives the second derivative $f''(x)$, differentiating again gives $f'''(x)$, and so on.
高阶导数就是导数的导数。对 $f'(x)$ 求导得到二阶导数（second derivative）$f''(x)$，再求一次导得到 $f'''(x)$，依此类推。
Notation
记号
Higher-order derivatives can be written in several equivalent ways:
高阶导数可以用以下几种等价的写法：
| Order阶数 | Prime Notation撇号记法 | Leibniz Notation莱布尼茨记法 |
|---|---|---|
| First一阶 | $f'(x)$ or $y'$$f'(x)$ 或 $y'$ | $\dfrac{dy}{dx}$ |
| Second二阶 | $f''(x)$ or $y''$$f''(x)$ 或 $y''$ | $\dfrac{d^2y}{dx^2}$ |
| Third三阶 | $f'''(x)$ or $y'''$$f'''(x)$ 或 $y'''$ | $\dfrac{d^3y}{dx^3}$ |
| $n$th$n$ 阶 | $f^{(n)}(x)$ | $\dfrac{d^ny}{dx^n}$ |
The process of finding higher-order derivatives mirrors the original differentiation process. "Function is to first derivative as first derivative is to second derivative." There is no new rule — you apply the same techniques repeatedly.
求高阶导数的过程与原本的求导过程完全一样。"函数与一阶导数的关系，正如一阶导数与二阶导数的关系。"没有新的法则——只是反复使用相同的技巧而已。
Find $f''(x)$ if $f(x) = \sin(2x)$.
已知 $f(x) = \sin(2x)$，求 $f''(x)$。
Observe how each successive derivative of $\sin(ax)$ picks up another factor of $a$. Use the order slider to reveal the pattern one derivative at a time.
观察 $\sin(ax)$ 的每一次连续求导都会多出一个因子 $a$。用阶数滑块逐阶揭示这一规律。
Implicit Second Derivatives
隐函数的二阶导数
To find $\frac{d^2 y}{dx^2}$ implicitly, differentiate $\frac{dy}{dx}$ (which itself contains $y$ and possibly $\frac{dy}{dx}$) with respect to $x$. Then substitute your expression for $\frac{dy}{dx}$ back in to express the result in terms of $x$ and $y$ only.
要用隐函数法求 $\frac{d^2 y}{dx^2}$，就对 $\frac{dy}{dx}$（其本身含有 $y$，可能还含 $\frac{dy}{dx}$）再对 $x$ 求导。然后把已经求好的 $\frac{dy}{dx}$ 的表达式代回，使最终结果只用 $x$ 和 $y$ 表示。
When differentiating $\frac{dy}{dx}$ implicitly, remember that $y$ is still a function of $x$. You will need the chain rule (and often the quotient rule) in this step.
用隐函数法对 $\frac{dy}{dx}$ 求导时，要记住 $y$ 仍是 $x$ 的函数。这一步会用到链式法则（常常还要用到商的法则）。
Find $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$ if $x^2 + y^2 = 25$.
已知 $x^2 + y^2 = 25$，求用 $x$、$y$ 表示的 $\dfrac{d^2y}{dx^2}$。
Step 1 — First derivative, implicitly.
第 1 步——先隐式求一阶导数。
$$2x + 2y\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{dy}{dx} = -\frac{x}{y}$$Step 2 — Differentiate again, using the quotient rule (and the chain rule on the $y$ in the denominator).
第 2 步——再求一次导，使用商的法则（分母中的 $y$ 还要用链式法则）。
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{(1)(y) - x\left(\frac{dy}{dx}\right)}{y^2}$$Step 3 — Substitute $\frac{dy}{dx} = -\frac{x}{y}$ back in and simplify. This is the step students most often skip — leaving $\frac{dy}{dx}$ in the answer is an incomplete simplification.
第 3 步——把 $\frac{dy}{dx} = -\frac{x}{y}$ 代回并化简。这一步是学生最容易漏掉的——答案中留着 $\frac{dy}{dx}$ 属于化简不完整。
$$\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}$$Step 4 — Use the original equation to simplify further. Since $x^2 + y^2 = 25$:
第 4 步——用原方程进一步化简。因为 $x^2 + y^2 = 25$：
$$\frac{d^2y}{dx^2} = -\frac{25}{y^3}$$This final substitution — recognizing that the constraint equation itself can simplify the answer — is a hallmark of a fully-worked implicit second-derivative problem and is exactly what separates a 9/9 from a partial-credit response on the free-response section.
这最后一步代入——认识到约束方程本身也能用来化简答案——正是完整解答隐函数二阶导数题目的标志，也是自由作答题中拿满分与只拿部分分的关键区别所在。
If $s(t)$ is position, $s'(t) = v(t)$ is velocity, and $s''(t) = a(t)$ is acceleration — a physical meaning for a second derivative you already know from Unit 2's rate-of-change work. The third derivative, $s'''(t)$, is called jerk: the rate at which acceleration itself changes. A smooth car ride has low jerk (acceleration changes gradually) even if the acceleration itself is nonzero; a jarring ride has high jerk. This gives every order of derivative a concrete interpretation, not just an abstract "derivative of a derivative."
若 $s(t)$ 是位置，则 $s'(t) = v(t)$ 是速度，$s''(t) = a(t)$ 是加速度——这是你在第 2 单元变化率内容中已经了解的二阶导数的物理意义。三阶导数 $s'''(t)$ 称为加加速度（jerk）：加速度本身变化的速率。乘坐平稳的汽车加加速度很低（加速度变化缓慢），即使加速度本身不为零；颠簸的乘坐体验则加加速度很高。这让每一阶导数都有了具体的解释，而不只是"导数的导数"这样一个抽象说法。
Exam Strategy
考试策略
Unit 3 content accounts for 9-13% of the AP Calculus AB exam and 4-7% of the BC exam. However, the chain rule is a foundational skill that appears indirectly in many questions across other units (related rates, optimization, integration by substitution, differential equations, and series). Mastering it here pays dividends throughout the course.
单元 3 的内容占 AP 微积分 AB 考试的 9–13%、BC 考试的 4–7%。但链式法则是一项基础技能，它会以间接方式出现在许多其他单元的题目里（相关变化率、最优化、换元积分、微分方程、级数）。在这里把它学扎实，整门课程都会受益。
Read the entire problem first. Determine which rules are needed before you start differentiating.
Show every application of the chain rule explicitly. On free-response questions, skipping the inner derivative is a guaranteed point deduction.
Practice with functions named $h$, $k$, $p$, etc. The AP Exam deliberately uses unfamiliar function names. Being comfortable with generic notation is essential.
Tables and graphs are common. Be ready to find derivatives of compositions and inverses using tabulated values — without an explicit formula.
先把整道题读完。在动笔求导之前先确定要用哪些法则。
每一次使用链式法则都要明确写出来。在简答题（free-response）上，跳过内函数的导数是一定要扣分的。
多用 $h$、$k$、$p$ 等函数名练习。AP 考试故意使用不熟悉的函数名。熟悉抽象记号是必不可少的能力。
表格和图像很常见。要做好准备，用表格中给出的数值——而不是显式公式——来求复合函数和反函数的导数。
When justifying answers, write the differentiation rule you are using. For example: "By the chain rule, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$." This demonstrates understanding and earns methodology points.
在说明答案时，写出所使用的求导法则。例如："由链式法则，$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$。"这能展示出你对概念的理解，并拿到方法分。
By the end of this unit you have six named techniques (power, product, quotient, chain, implicit, inverse — plus logarithmic differentiation as a bonus tool). The AP Exam rarely tests one in isolation on the harder questions; it tests whether you can read a nested expression, identify every layer, and apply the right tool at each layer in the right order. When you get stuck, don't search for "the one rule" — ask "what's the outermost structure, and what's inside it?" and work inward one layer at a time.
学完本单元后，你已经掌握了六种命名法则（幂、乘积、商、链式、隐函数、反函数——外加对数求导法这个额外工具）。AP 考试中较难的题目很少只单独考查一条法则；它考查的是你能否读懂一个嵌套表达式、识别出每一层结构，并按正确顺序在每一层使用恰当的工具。卡壳时，不要去找"唯一那一条法则"，而要问"最外层结构是什么？里面又是什么？"，然后一层一层向内处理。
Common Mistakes
常见错误
The most frequent error: writing $\frac{d}{dx}[\sin(3x)] = \cos(3x)$ instead of $3\cos(3x)$. Always multiply by $g'(x)$.
最常见的错误：把 $\frac{d}{dx}[\sin(3x)]$ 写成 $\cos(3x)$，而不是 $3\cos(3x)$。永远要乘以 $g'(x)$。
In implicit differentiation, every differentiated $y$ term must carry a $\frac{dy}{dx}$ factor. Writing $\frac{d}{dx}[y^3] = 3y^2$ instead of $3y^2 \frac{dy}{dx}$ is incorrect.
在隐函数求导中，每一个被求导的 $y$ 项都必须带上一个因子 $\frac{dy}{dx}$。把 $\frac{d}{dx}[y^3]$ 写成 $3y^2$ 而不是 $3y^2 \frac{dy}{dx}$ 是错的。
These have different outer/inner structures. $(\sin x)^4$ has outer $u^4$ and inner $\sin x$. But $\sin(x^4)$ has outer $\sin u$ and inner $x^4$. Misidentifying gives a wrong answer.
这两种写法的外/内函数结构不同。$(\sin x)^4$ 的外函数是 $u^4$，内函数是 $\sin x$；而 $\sin(x^4)$ 的外函数是 $\sin u$，内函数是 $x^4$。识别错误就会得到错误答案。
When using $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a)=b$, students sometimes evaluate $f'$ at $b$ instead of at $a$. Remember: evaluate the original function's derivative at the input that maps to $b$.
使用 $(f^{-1})'(b) = \frac{1}{f'(a)}$ 时（其中 $f(a)=b$），学生有时会把 $f'$ 代到 $b$ 处而不是 $a$ 处。要记住：原函数的导数应代入那个映射到 $b$ 的自变量值。
Students sometimes apply the chain rule to the outermost layer but forget that $e^{g(x)}$ still needs $g'(x)$, or that $\cos(g(x))$ needs $g'(x)$. Carry the chain through every layer.
学生有时只对最外层用了链式法则，却忘了 $e^{g(x)}$ 还要乘 $g'(x)$，或 $\cos(g(x))$ 也要乘 $g'(x)$。要把链式贯穿到每一层。
After differentiating $\ln y$ implicitly, you get $\frac{1}{y}\frac{dy}{dx} = (\text{something})$. The final answer needs $\frac{dy}{dx}$ alone, so you must multiply both sides by $y$ — and then substitute the original expression for $y$ back in. Leaving a bare $y$ (or worse, stopping at $\frac{1}{y}\frac{dy}{dx} = \ldots$) is an incomplete answer.
对 $\ln y$ 隐式求导后，会得到 $\frac{1}{y}\frac{dy}{dx} = (\text{某表达式})$。最终答案需要单独的 $\frac{dy}{dx}$，所以必须两边同乘 $y$——然后把 $y$ 的原始表达式代回去。留下裸露的 $y$（或者更糟，停在 $\frac{1}{y}\frac{dy}{dx} = \ldots$）都是不完整的答案。
Flashcards
闪卡
Click any card to reveal the answer. Click again to flip back.
点击任意一张卡查看答案。再次点击翻回正面。
given $f(a) = b$
已知 $f(a) = b$
Unit Quiz
单元小测
Test your understanding. Select an answer for each question, then check the explanation.
检验你的理解。为每道题选择一个答案，再查看解释。
Readiness Checklist
备考清单
Click each item you've mastered. Aim for 100% before exam day.
点击每一条你已经掌握的内容。目标是在考试前达到 100%。
- Apply the Chain Rule to nested composite functions对嵌套的复合函数使用链式法则
- Differentiate triple-composed functions without losing a factor对三层复合的函数求导时不遗漏任何一个因子
- Combine the Chain Rule with Product and Quotient Rules把链式法则与乘积法则、商的法则结合使用
- Perform implicit differentiation to find $dy/dx$使用隐函数求导求 $dy/dx$
- Write tangent line equations to implicitly defined curves写出隐式定义曲线的切线方程
- Differentiate an inverse function using $(f^{-1})'(a) = 1/f'(f^{-1}(a))$用 $(f^{-1})'(a) = 1/f'(f^{-1}(a))$ 对反函数求导
- Differentiate $\arcsin$, $\arccos$, $\arctan$ and related expressions对 $\arcsin$、$\arccos$、$\arctan$ 及相关表达式求导
- Compute higher-order derivatives $f''$, $f'''$, $f^{(n)}$计算高阶导数 $f''$、$f'''$、$f^{(n)}$
- Find $d^2y/dx^2$ implicitly in terms of $x$ and $y$用隐函数法求 $d^2y/dx^2$，结果用 $x$ 和 $y$ 表示
- Select the correct differentiation procedure for a given expression为给定表达式选择正确的求导方法
- Recognize when logarithmic differentiation simplifies the work识别何时使用对数求导法（
logarithmic differentiation）能简化运算 - Simplify derivative expressions and factor out common terms化简导数表达式并提取公因子
- Use the Chain Rule in Leibniz form $dy/dx = (dy/du)(du/dx)$用莱布尼茨形式的链式法则 $dy/dx = (dy/du)(du/dx)$
- Interpret $f''$ qualitatively (concavity, acceleration)从定性角度解读 $f''$（凹凸性、加速度）
AP-Style Practice ProblemsAP 风格练习题
Exam-level practice for this unit — multiple-choice plus extended-response items modeled on the AP rubric. Built for top-score prep; go here after you've worked through the notes and the in-page quizzes above.
本单元的考试级练习——按 AP 评分标准设计的多项选择题与简答题。为冲击高分而准备；学完上方笔记和页内小测后再做。