Unit A4: Derivatives of
Transcendental Functions单元 A4：超越函数的导数

Apply the limit definition and expand with the identity $\sin(x+h)=\sin x\cos h+\cos x\sin h$:

Group the terms that carry $\sin x$ and those that carry $\cos x$:

The two parenthesized limits are exactly the fundamental trig limits. The same procedure with $\cos(x+h)$ yields $\frac{d}{dx}\cos x=-\sin x$.

用导数的极限定义，并借助恒等式 $\sin(x+h)=\sin x\cos h+\cos x\sin h$ 展开：

把带 $\sin x$ 的项和带 $\cos x$ 的项各自归并：

括号里的两个极限正是前面那两个基本三角极限。对 $\cos(x+h)$ 做同样的处理就得到 $\frac{d}{dx}\cos x=-\sin x$。

This single limit is the foundation of every formula in this section, so it deserves a complete argument rather than an appeal to a picture. Fix a small angle $x$ with $0 $$ \underbrace{\tfrac12\sin x}_{\text{inner triangle}} \;\le\; \underbrace{\tfrac12 x}_{\text{sector}} \;\le\; \underbrace{\tfrac12\tan x}_{\text{outer triangle}}. $$

The sector area is $\tfrac12 r^2 x=\tfrac12 x$ precisely because $r=1$ and $x$ is in radians; this is the step that fails in degrees. Multiply through by $\dfrac{2}{\sin x}>0$:

Take reciprocals, which reverses the inequalities:

As $x\to 0^+$ we have $\cos x\to 1$, so the squeeze theorem forces $\frac{\sin x}{x}\to 1$. Because $\frac{\sin x}{x}$ is even, the same limit holds as $x\to 0^-$, hence as $x\to 0$. The companion limit follows by the conjugate trick: $\frac{1-\cos x}{x}=\frac{1-\cos^2 x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\to 1\cdot\frac{0}{2}=0$.

这一个极限是本节每条公式的地基，所以值得给出完整论证，而不是诉诸一张图。取一个满足 $0 $$ \underbrace{\tfrac12\sin x}_{\text{inner triangle}} \;\le\; \underbrace{\tfrac12 x}_{\text{sector}} \;\le\; \underbrace{\tfrac12\tan x}_{\text{outer triangle}}. $$

扇形面积之所以是 $\tfrac12 r^2 x=\tfrac12 x$，正因为 $r=1$ 且 $x$ 以弧度计；这一步在角度制下会失效。两边同乘 $\dfrac{2}{\sin x}>0$：

取倒数，不等号方向随之翻转：

当 $x\to 0^+$ 时 $\cos x\to 1$，于是夹逼定理迫使 $\frac{\sin x}{x}\to 1$。由于 $\frac{\sin x}{x}$ 是偶函数，$x\to 0^-$ 时极限相同，因而 $x\to 0$ 时也成立。配套极限由共轭技巧得到：$\frac{1-\cos x}{x}=\frac{1-\cos^2 x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\to 1\cdot\frac{0}{2}=0$。

Differentiate $g(x) = 3\sin x - 5\cos x$.

The constants ride through unchanged by linearity, and each term follows its own rule.

对 $g(x) = 3\sin x - 5\cos x$ 求导。

由线性性质，常数系数原样保留，每一项各按自己的法则求导。

Differentiate $f(x)=x^2\cos(3x)$. This is a product, and the second factor is a composition, so both the product rule and the chain rule appear. By the product rule,

The first derivative is $2x$. For the second, the outer function is cosine and the inner is $3x$ with inner derivative $3$, so $\frac{d}{dx}\cos(3x)=-\sin(3x)\cdot 3=-3\sin(3x)$. Assembling,

Notice the inner derivative $3$ is not optional: dropping it is the single most common error on composite trig derivatives.

对 $f(x)=x^2\cos(3x)$ 求导。这是一个乘积，且第二个因子是复合函数，所以乘积法则（Product Rule）和链式法则（Chain Rule）都要出场。由乘积法则，

第一个导数是 $2x$。第二个里，外层是余弦，内层是 $3x$，其内层导数为 $3$，所以 $\frac{d}{dx}\cos(3x)=-\sin(3x)\cdot 3=-3\sin(3x)$。合并：

注意内层导数 $3$ 不可省略：漏掉它是复合三角函数求导中最常见的错误。

Find the slope of the tangent line to $y=\sin x$ at $x=\tfrac{\pi}{3}$, and confirm where the curve is momentarily flat on $[0,2\pi]$.

The derivative is $y'=\cos x$, so the slope at $x=\tfrac{\pi}{3}$ is $\cos\tfrac{\pi}{3}=\tfrac12$. The tangent rises at a moderate rate, consistent with the graph still climbing toward its peak at $\tfrac{\pi}{2}$.

The curve is flat where $y'=0$, that is where $\cos x=0$. On $[0,2\pi]$ that happens at $x=\tfrac{\pi}{2}$ and $x=\tfrac{3\pi}{2}$, exactly the maximum and the minimum of the sine wave. The derivative formula thus locates the turning points without any further work.

求 $y=\sin x$ 在 $x=\tfrac{\pi}{3}$ 处切线的斜率，并确认曲线在 $[0,2\pi]$ 上何处瞬时平坦。

导数是 $y'=\cos x$，所以在 $x=\tfrac{\pi}{3}$ 处斜率为 $\cos\tfrac{\pi}{3}=\tfrac12$。切线以适中的速度上升，与图像仍在向 $\tfrac{\pi}{2}$ 处的峰值攀升一致。

曲线在 $y'=0$ 处平坦，即 $\cos x=0$ 处。在 $[0,2\pi]$ 上这发生在 $x=\tfrac{\pi}{2}$ 和 $x=\tfrac{3\pi}{2}$，恰好是正弦波的最大值与最小值。可见，导数公式无需额外功夫就定位了这些转折点。

Write $\tan x=\dfrac{\sin x}{\cos x}$ and apply the quotient rule:

The numerator is the Pythagorean identity $\cos^2 x+\sin^2 x=1$, so

把 $\tan x=\dfrac{\sin x}{\cos x}$ 写出来，套用商的法则：

分子是毕达哥拉斯恒等式 $\cos^2 x+\sin^2 x=1$，于是

Write $\sec x=(\cos x)^{-1}$ and use the chain rule on the power. With outer function $u^{-1}$ and inner $u=\cos x$, $u'=-\sin x$,

Now split the single fraction into a product that exposes the standard form:

The same method on $\csc x=(\sin x)^{-1}$ gives $\frac{d}{dx}\csc x=-\csc x\cot x$, and on $\cot x=(\tan x)^{-1}$, or directly as $\frac{\cos x}{\sin x}$ by the quotient rule, gives $\frac{d}{dx}\cot x=-\csc^2 x$.

把 $\sec x=(\cos x)^{-1}$ 写出来，对这个幂用链式法则。外层是 $u^{-1}$，内层 $u=\cos x$，$u'=-\sin x$，

再把这个单一分式拆成乘积，露出标准形式：

对 $\csc x=(\sin x)^{-1}$ 用同样方法得 $\frac{d}{dx}\csc x=-\csc x\cot x$；对 $\cot x=(\tan x)^{-1}$（或直接写成 $\frac{\cos x}{\sin x}$ 用商的法则）得 $\frac{d}{dx}\cot x=-\csc^2 x$。

Differentiate $h(x) = x\sec x$ using the product rule.

用乘积法则对 $h(x) = x\sec x$ 求导。

Differentiate $f(x)=\dfrac{\tan x}{x}$. By the quotient rule with top $\tan x$ and bottom $x$,

There is no further cancellation, so this is the final simplified form. Resist the temptation to cancel an $x$ against the $x^2$, since the second term in the numerator has no factor of $x$ to cancel.

对 $f(x)=\dfrac{\tan x}{x}$ 求导。用商的法则，分子为 $\tan x$、分母为 $x$，

没有进一步可约的因子，所以这就是最简形式。不要被诱惑去拿 $x$ 跟 $x^2$ 约分，因为分子第二项没有 $x$ 因子可约。

Differentiate $g(x)=\sec^3 x$, which means $(\sec x)^3$. The outer operation is the cube, so the general power rule applies with inner function $\sec x$:

The chain rule supplies the inner derivative $\sec x\tan x$, which then merges with the $\sec^2 x$ from the power rule to give $\sec^3 x$.

对 $g(x)=\sec^3 x$ 求导，它的含义是 $(\sec x)^3$。外层运算是立方，所以对内层函数 $\sec x$ 套用广义幂法则：

链式法则提供内层导数 $\sec x\tan x$，它再与幂法则给出的 $\sec^2 x$ 合并成 $\sec^3 x$。

From the definition, factor $a^x$ out of the difference quotient for $f(x)=a^x$:

The remaining limit is a constant $L(a)$ that depends only on the base. The number $e$ is defined as the base for which this limit equals $1$, that is $\lim_{h\to 0}\frac{e^h-1}{h}=1$. Therefore $\frac{d}{dx}e^x=e^x\cdot 1=e^x$. For general $a$ one shows $L(a)=\ln a$, giving $\frac{d}{dx}a^x=a^x\ln a$.

从定义出发，对 $f(x)=a^x$ 的差商把 $a^x$ 提出来：

余下的极限（limit）是一个只依赖于底数的常数 $L(a)$。数 $e$ 被定义为使该极限等于 $1$ 的那个底数，即 $\lim_{h\to 0}\frac{e^h-1}{h}=1$。因此 $\frac{d}{dx}e^x=e^x\cdot 1=e^x$。对一般的 $a$ 可证 $L(a)=\ln a$，从而 $\frac{d}{dx}a^x=a^x\ln a$。

The proof above left $L(a)=\lim_{h\to 0}\frac{a^h-1}{h}$ as an unidentified constant. Here is why it equals $\ln a$, using only the base-$e$ rule. Write $a^x=e^{x\ln a}$, a genuine identity since $a=e^{\ln a}$. Differentiate the right side with the chain rule, outer function $e^u$ and inner $u=x\ln a$ with $u'=\ln a$:

Comparing this with the difference-quotient form $\frac{d}{dx}a^x=a^x\,L(a)$ and matching coefficients of $a^x$ forces $L(a)=\ln a$. As a sanity check, $L(e)=\ln e=1$, recovering the defining property of $e$, and for a base less than $1$, say $a=\tfrac12$, $L(a)=\ln\tfrac12<0$, correctly making $(\tfrac12)^x$ a decreasing function.

上面的证明把 $L(a)=\lim_{h\to 0}\frac{a^h-1}{h}$ 留作一个未辨认的常数。下面只用以 $e$ 为底的法则说明它为何等于 $\ln a$。写出 $a^x=e^{x\ln a}$，这是真正的恒等式，因为 $a=e^{\ln a}$。对右边用链式法则求导，外层是 $e^u$，内层 $u=x\ln a$ 且 $u'=\ln a$：

把它与差商形式 $\frac{d}{dx}a^x=a^x\,L(a)$ 比较，匹配 $a^x$ 的系数就迫使 $L(a)=\ln a$。作为验算，$L(e)=\ln e=1$，恰好还原 $e$ 的定义性质；而对小于 $1$ 的底数，比如 $a=\tfrac12$，$L(a)=\ln\tfrac12<0$，正确地使 $(\tfrac12)^x$ 成为减函数。

Differentiate $f(x) = e^{3x^2}$. Here $u=3x^2$, so $u'=6x$:

对 $f(x) = e^{3x^2}$ 求导。这里 $u=3x^2$，所以 $u'=6x$：

Differentiate $f(x)=\dfrac{e^{x}}{x^2+1}$. By the quotient rule, with top $e^x$ (derivative $e^x$) and bottom $x^2+1$ (derivative $2x$),

The numerator factors as a perfect square, $x^2-2x+1=(x-1)^2$, so

Because $(x-1)^2\ge 0$ and $e^x>0$, the derivative is never negative: $f$ is increasing everywhere, with a momentary flat spot at $x=1$ where the tangent is horizontal.

对 $f(x)=\dfrac{e^{x}}{x^2+1}$ 求导。用商的法则，分子为 $e^x$（导数 $e^x$）、分母为 $x^2+1$（导数 $2x$），

分子是完全平方 $x^2-2x+1=(x-1)^2$，于是

因为 $(x-1)^2\ge 0$ 且 $e^x>0$，导数从不为负：$f$ 处处递增，仅在 $x=1$ 处有一个瞬时平点，那里切线水平。

Differentiate $y=5^{\,\sin x}$. The base is a constant and the exponent is a function, so rewrite in base $e$: $5^{\sin x}=e^{(\sin x)\ln 5}$. Now the chain rule applies cleanly with inner function $(\sin x)\ln 5$ and inner derivative $(\cos x)\ln 5$:

Equivalently, apply $\frac{d}{dx}a^{u}=a^{u}(\ln a)\,u'$ directly with $a=5$ and $u=\sin x$. Both routes agree, which is the point of the conversion identity.

对 $y=5^{\,\sin x}$ 求导。底数是常数、指数是函数，于是换成以 $e$ 为底：$5^{\sin x}=e^{(\sin x)\ln 5}$。这时链式法则可干净地应用，内层函数为 $(\sin x)\ln 5$，内层导数为 $(\cos x)\ln 5$：

等价地，直接用 $\frac{d}{dx}a^{u}=a^{u}(\ln a)\,u'$，取 $a=5$、$u=\sin x$。两条路径结果一致，这正是换底恒等式的意义所在。

Let $y=\ln x$, so $e^{y}=x$. Differentiate both sides with respect to $x$, using the chain rule on the left:

The substitution $e^{y}=x$ closes the argument and gives the reciprocal formula.

令 $y=\ln x$，于是 $e^{y}=x$。两边对 $x$ 求导，左边用链式法则：

代回 $e^{y}=x$ 就完成论证，给出倒数公式。

The natural logarithm is special because its derivative has no leftover constant. Every other base inherits a factor of $\frac{1}{\ln a}$ through the change-of-base identity. Start from

which is valid because $\ln a$ is just a constant. Differentiate, treating $\frac{1}{\ln a}$ as a constant multiple:

This also explains why calculus uses base $e$ almost exclusively: it is the one base for which the constant $\frac{1}{\ln a}$ equals $1$, leaving the cleanest possible derivative.

自然对数之所以特殊，是因为它的导数没有多余的常数。其它每一个底数都通过换底恒等式继承一个 $\frac{1}{\ln a}$ 因子。从

出发，这成立是因为 $\ln a$ 只是个常数。把 $\frac{1}{\ln a}$ 当成常数倍求导：

这也解释了为什么微积分几乎只用以 $e$ 为底：它是唯一让常数 $\frac{1}{\ln a}$ 等于 $1$ 的底数，留下尽可能简洁的导数。

Differentiate $y = x^{x}$ for $x>0$. The variable appears in both base and exponent, so neither the power rule nor the exponential rule applies directly. Take logarithms:

Differentiate both sides, using the product rule on the right:

Multiply by $y=x^{x}$:

对 $x>0$ 求 $y = x^{x}$ 的导数。变量同时出现在底数和指数上，所以幂法则和指数法则都不能直接用。两边取对数：

两边求导，右边用乘积法则（Product Rule）：

乘以 $y=x^{x}$：

Differentiate $y=\dfrac{(x^2+1)^3\sqrt{x-4}}{(2x+3)^5}$ for $x>4$. The product and quotient rules would be brutal here. Take the natural log and use its algebra to break the expression into a sum:

Now differentiate term by term, each by the chain rule on $\ln$:

Multiply by $y$ to finish:

The bracket is the logarithmic derivative; leaving the answer in this factored form is standard and preferred, since expanding it gains nothing.

对 $x>4$ 求 $y=\dfrac{(x^2+1)^3\sqrt{x-4}}{(2x+3)^5}$ 的导数。在这里乘积法则和商的法则会异常繁琐。取自然对数，用它的运算律把式子拆成和：

现在逐项求导，每一项都对 $\ln$ 用链式法则：

乘以 $y$ 收尾：

方括号里就是对数导数；把答案保留成这种因式形式是标准且更可取的做法，因为展开它没有任何好处。

Differentiate $y=(\sin x)^{x}$ on an interval where $\sin x>0$, say $0 $$ \ln y = x\ln(\sin x). $$

Differentiate the right side by the product rule, and the inner $\ln(\sin x)$ by the chain rule:

Multiply by $y=(\sin x)^x$:

在 $\sin x>0$ 的区间上（比如 $0 $$ \ln y = x\ln(\sin x). $$

右边用乘积法则求导，内层 $\ln(\sin x)$ 用链式法则：

乘以 $y=(\sin x)^x$：

Let $y=\arcsin x$, so $\sin y=x$ with $-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}$. Differentiate implicitly:

On the principal branch $\cos y\ge 0$, so $\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2}$, giving

令 $y=\arcsin x$，于是 $\sin y=x$，其中 $-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}$。做隐函数求导：

在主支上 $\cos y\ge 0$，所以 $\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2}$，于是

Let $y=\arctan x$, so $\tan y=x$ with $-\tfrac{\pi}{2} $$ \sec^2 y\,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\sec^2 y}. $$

Now convert the leftover trig to algebra with the identity $\sec^2 y=1+\tan^2 y=1+x^2$:

Here no sign ambiguity arises because $\sec^2 y$ is always positive, which is why arctangent has no branch-sign subtlety. Contrast this with $\arcsin$, where choosing $\cos y=+\sqrt{1-x^2}$ rather than the negative root depended on the principal branch lying in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, where cosine is nonnegative. Choosing the wrong root there flips the sign of the entire derivative, the most common error in this section.

令 $y=\arctan x$，于是 $\tan y=x$，其中 $-\tfrac{\pi}{2} $$ \sec^2 y\,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\sec^2 y}. $$

再用恒等式 $\sec^2 y=1+\tan^2 y=1+x^2$ 把残留的三角化成代数：

这里不存在符号歧义，因为 $\sec^2 y$ 永远为正，这正是反正切没有分支符号难点的原因。与之对照的是 $\arcsin$：选取 $\cos y=+\sqrt{1-x^2}$ 而非负根，依赖于主支落在 $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ 内，那里余弦非负。在那里选错根会翻转整个导数的符号，这是本节最常见的错误。

Differentiate $f(x) = \arctan(x^2)$. With $u=x^2$ and $u'=2x$:

对 $f(x) = \arctan(x^2)$ 求导。取 $u=x^2$、$u'=2x$：

Differentiate $g(x)=\arcsin(3x)$. The outer function is $\arcsin u$ with $u=3x$ and $u'=3$:

The domain shrinks accordingly: the radicand $1-9x^2$ stays positive only for $|x|<\tfrac13$, exactly where $3x$ lands inside $(-1,1)$.

对 $g(x)=\arcsin(3x)$ 求导。外层函数是 $\arcsin u$，其中 $u=3x$、$u'=3$：

定义域随之收窄：被开方式 $1-9x^2$ 仅在 $|x|<\tfrac13$ 时为正，正是 $3x$ 落入 $(-1,1)$ 的地方。

Differentiate $h(x)=x\arctan x - \tfrac12\ln(1+x^2)$, a function that appears as an antiderivative in integration. Differentiate term by term. For the first term use the product rule:

For the second, the chain rule on $\ln(1+x^2)$ gives $\frac{2x}{1+x^2}$, so $\frac{d}{dx}\big(-\tfrac12\ln(1+x^2)\big)=-\tfrac12\cdot\frac{2x}{1+x^2}=-\frac{x}{1+x^2}$. Adding the pieces, the two rational terms cancel:

The clean cancellation is the reason $h$ is the standard antiderivative of $\arctan x$.

对 $h(x)=x\arctan x - \tfrac12\ln(1+x^2)$ 求导，这个函数在积分里作为 $\arctan x$ 的原函数（antiderivative）出现。逐项求导。第一项用乘积法则：

第二项对 $\ln(1+x^2)$ 用链式法则得 $\frac{2x}{1+x^2}$，于是 $\frac{d}{dx}\big(-\tfrac12\ln(1+x^2)\big)=-\tfrac12\cdot\frac{2x}{1+x^2}=-\frac{x}{1+x^2}$。把各部分相加，两个有理项相消：

这种干净的相消正是 $h$ 成为 $\arctan x$ 标准原函数的原因。

Differentiate the exponential definition term by term:

No minus sign survives, in contrast to the circular case where $\frac{d}{dx}\cos x=-\sin x$.

对指数定义逐项求导：

没有负号留存，这与圆三角情形 $\frac{d}{dx}\cos x=-\sin x$ 形成对照。

Write $\tanh x=\dfrac{\sinh x}{\cosh x}$ and apply the quotient rule, using $\frac{d}{dx}\sinh x=\cosh x$ and $\frac{d}{dx}\cosh x=\sinh x$:

The numerator is the hyperbolic Pythagorean identity $\cosh^2 x-\sinh^2 x=1$, which itself follows from the definitions since $\big(\tfrac{e^x+e^{-x}}{2}\big)^2-\big(\tfrac{e^x-e^{-x}}{2}\big)^2=\tfrac14\big[(e^{2x}+2+e^{-2x})-(e^{2x}-2+e^{-2x})\big]=1$. Therefore

Note the sign of the numerator: in the circular case the quotient rule produced $\cos^2 x+\sin^2 x$, a sum, whereas here the minus signs in the hyperbolic derivatives produce a difference. Both collapse to $1$ by their respective identities.

把 $\tanh x=\dfrac{\sinh x}{\cosh x}$ 写出来，套用商的法则，用 $\frac{d}{dx}\sinh x=\cosh x$ 和 $\frac{d}{dx}\cosh x=\sinh x$：

分子是双曲毕达哥拉斯恒等式 $\cosh^2 x-\sinh^2 x=1$，它本身由定义而来，因为 $\big(\tfrac{e^x+e^{-x}}{2}\big)^2-\big(\tfrac{e^x-e^{-x}}{2}\big)^2=\tfrac14\big[(e^{2x}+2+e^{-2x})-(e^{2x}-2+e^{-2x})\big]=1$。因此

注意分子的符号：在圆三角情形里商的法则产生 $\cos^2 x+\sin^2 x$，是一个和；而这里双曲导数里的负号产生一个差。两者都按各自的恒等式化为 $1$。

Differentiate $f(x) = \cosh(2x)$. With $u=2x$ and $u'=2$:

对 $f(x) = \cosh(2x)$ 求导。取 $u=2x$、$u'=2$：

A hanging cable, the catenary, has shape $y=a\cosh\!\big(\tfrac{x}{a}\big)$ for a positive constant $a$. Find its slope. The outer function is $a\cosh u$ with $u=\tfrac{x}{a}$ and $u'=\tfrac1a$:

The constant $a$ cancels, so the slope at the lowest point $x=0$ is $\sinh 0=0$, confirming the cable hangs horizontally at its bottom. The arc length element $\sqrt{1+(y')^2}=\sqrt{1+\sinh^2(x/a)}=\cosh(x/a)$ then simplifies beautifully, which is exactly why the catenary is the standard textbook arc-length example.

一根悬挂的缆绳——悬链线——其形状为 $y=a\cosh\!\big(\tfrac{x}{a}\big)$，其中 $a$ 为正常数。求它的斜率。外层函数是 $a\cosh u$，其中 $u=\tfrac{x}{a}$、$u'=\tfrac1a$：

常数 $a$ 相消，所以在最低点 $x=0$ 处斜率为 $\sinh 0=0$，证实缆绳在底部水平悬垂。弧长微元 $\sqrt{1+(y')^2}=\sqrt{1+\sinh^2(x/a)}=\cosh(x/a)$ 随之漂亮地化简，这正是悬链线成为教科书标准弧长例子的原因。

Differentiate $g(x)=\tanh(x^2+1)$. The outer function is $\tanh u$ with derivative $\operatorname{sech}^2 u$, and the inner is $u=x^2+1$ with $u'=2x$:

对 $g(x)=\tanh(x^2+1)$ 求导。外层函数是 $\tanh u$，导数为 $\operatorname{sech}^2 u$；内层是 $u=x^2+1$、$u'=2x$：

It is worth seeing once why the running product is the whole story for a deep composition. Suppose $F=f_1\circ f_2\circ\cdots\circ f_n$, so $F(x)=f_1\big(f_2(\cdots f_n(x))\big)$. Let $u_n=f_n(x)$, $u_{n-1}=f_{n-1}(u_n)$, and so on. Applying the two-function chain rule once peels the outermost layer:

The trailing factor is again a composition, one layer shorter, so the same rule applies to it. Repeating until nothing is left gives the telescoping product

which is exactly the running product of inner derivatives, each evaluated at the appropriate intermediate value. Forgetting one of these factors is forgetting one application of the chain rule, which is why the layer-by-layer habit is not optional.

值得彻底看一次：对深层复合而言，累乘内层导数就是全部故事。设 $F=f_1\circ f_2\circ\cdots\circ f_n$，即 $F(x)=f_1\big(f_2(\cdots f_n(x))\big)$。令 $u_n=f_n(x)$、$u_{n-1}=f_{n-1}(u_n)$，依此类推。对两个函数的链式法则用一次，就剥掉最外层：

末尾那个因子又是一个复合，只是少了一层，所以对它再用同一法则。反复直到无可再剥，得到这个望远镜式的连乘

这正是累乘的内层导数，每一个都在恰当的中间值处取值。漏掉其中一个因子，就是漏掉一次链式法则的应用，这正是逐层习惯不可省略的原因。

Differentiate $f(x) = e^{2x}\sin(3x)$. This is a product, and each factor needs the chain rule:

对 $f(x) = e^{2x}\sin(3x)$ 求导。这是一个乘积，每个因子都需要链式法则：

Differentiate $g(x) = \ln\!\big(\tan x\big)$. The outer function is $\ln u$ with $u=\tan x$:

对 $g(x) = \ln\!\big(\tan x\big)$ 求导。外层函数是 $\ln u$，其中 $u=\tan x$：

Differentiate $f(x)=\arctan\!\big(e^{\sin x}\big)$. Three operations are stacked: arctangent on the outside, exponential in the middle, sine on the inside. Differentiate outward in, keeping the running product.

Outer layer, arctangent, gives $\dfrac{1}{1+\big(e^{\sin x}\big)^2}=\dfrac{1}{1+e^{2\sin x}}$, with its argument left untouched. Middle layer, the exponential, contributes $e^{\sin x}$. Inner layer, sine, contributes $\cos x$. Multiply the three factors:

Each factor is the derivative of one named layer, and none is skipped. That is the entire technique.

对 $f(x)=\arctan\!\big(e^{\sin x}\big)$ 求导。三个运算叠在一起：外层反正切，中层指数，内层正弦。由外向内求导，一路累乘。

最外层反正切给出 $\dfrac{1}{1+\big(e^{\sin x}\big)^2}=\dfrac{1}{1+e^{2\sin x}}$，其自变量保持不动。中层指数贡献 $e^{\sin x}$。内层正弦贡献 $\cos x$。把三个因子相乘：

每个因子都是某个具名层的导数，无一被跳过。这就是这套技术的全部。

Differentiate $y=\dfrac{\sin x}{e^{x}}$, which is cleaner to read as $y=e^{-x}\sin x$. Using the product rule on the rewritten form, with $\frac{d}{dx}e^{-x}=-e^{-x}$,

This $e^{-x}\sin x$ is the prototype of a damped oscillation, and the factored derivative $e^{-x}(\cos x-\sin x)$ vanishes exactly when $\cos x=\sin x$, that is at $x=\tfrac{\pi}{4}+k\pi$, locating the peaks and troughs of the decaying wave.

对 $y=\dfrac{\sin x}{e^{x}}$ 求导，写成 $y=e^{-x}\sin x$ 更便于阅读。对改写后的形式用乘积法则，其中 $\frac{d}{dx}e^{-x}=-e^{-x}$，

这个 $e^{-x}\sin x$ 是阻尼振荡的原型，因式化后的导数 $e^{-x}(\cos x-\sin x)$ 恰好在 $\cos x=\sin x$ 时为零，即 $x=\tfrac{\pi}{4}+k\pi$，定位了这条衰减波的波峰与波谷。