Unit A5: Implicit Differentiation and Related Rates单元 A5：隐函数求导与相关变化率

Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$ and the slope of the tangent at $(3,4)$.

Differentiate both sides with respect to $x$:

$$ 2x + 2y\,\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{dy}{dx} = -\frac{x}{y}. $$

At $(3,4)$ the slope is $-\tfrac{3}{4}$. Note the answer depends on both coordinates, which is typical of implicit differentiation: a single $x$ may correspond to several points on the curve, each with its own slope.

对圆 $x^2 + y^2 = 25$ 求 $\frac{dy}{dx}$，并求在 $(3,4)$ 处切线的斜率。

对两边关于 $x$ 求导：

$$ 2x + 2y\,\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{dy}{dx} = -\frac{x}{y}. $$

在 $(3,4)$ 处斜率为 $-\tfrac{3}{4}$。注意答案同时依赖于两个坐标，这正是隐函数求导的典型特征：同一个 $x$ 可能对应曲线上的多个点，每个点各有自己的斜率。

Find $\frac{dy}{dx}$ for the folium of Descartes $x^3 + y^3 = 6xy$.

Differentiate term by term, using the product rule on $6xy$:

$$ 3x^2 + 3y^2\,\frac{dy}{dx} = 6\Big(y + x\,\frac{dy}{dx}\Big). $$

Collect the $\frac{dy}{dx}$ terms:

$$ \big(3y^2 - 6x\big)\frac{dy}{dx} = 6y - 3x^2 \;\Longrightarrow\; \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}. $$

At $(3,3)$ the slope is $\frac{6-9}{9-6} = -1$, so the tangent there is the line $y = -x + 6$.

对笛卡尔叶形线 $x^3 + y^3 = 6xy$ 求 $\frac{dy}{dx}$。

逐项求导，对 $6xy$ 使用乘积法则（Product Rule）：

$$ 3x^2 + 3y^2\,\frac{dy}{dx} = 6\Big(y + x\,\frac{dy}{dx}\Big). $$

归并含 $\frac{dy}{dx}$ 的项：

$$ \big(3y^2 - 6x\big)\frac{dy}{dx} = 6y - 3x^2 \;\Longrightarrow\; \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}. $$

在 $(3,3)$ 处斜率为 $\frac{6-9}{9-6} = -1$，所以该处切线为直线 $y = -x + 6$。

Find the equation of the tangent to the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at the point $\big(2,\,\tfrac{3\sqrt{3}}{2}\big)$.

Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

$$ \frac{2x}{16} + \frac{2y}{9}\,\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{x}{8} + \frac{2y}{9}\,\frac{dy}{dx} = 0. $$

Solve for the slope:

$$ \frac{dy}{dx} = -\frac{9x}{16y}. $$

At $\big(2,\,\tfrac{3\sqrt{3}}{2}\big)$ the slope is $-\dfrac{9(2)}{16\cdot\frac{3\sqrt 3}{2}} = -\dfrac{18}{24\sqrt 3} = -\dfrac{3}{4\sqrt 3} = -\dfrac{\sqrt 3}{4}$. The tangent line is therefore $y - \tfrac{3\sqrt 3}{2} = -\tfrac{\sqrt 3}{4}\,(x - 2)$. Solving explicitly for $y$ would force a square root and a choice of branch, so implicit differentiation is the cleaner route.

求椭圆 $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ 在点 $\big(2,\,\tfrac{3\sqrt{3}}{2}\big)$ 处的切线方程。

对两边关于 $x$ 求导，把 $y$ 当作 $x$ 的函数：

$$ \frac{2x}{16} + \frac{2y}{9}\,\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{x}{8} + \frac{2y}{9}\,\frac{dy}{dx} = 0. $$

解出斜率：

$$ \frac{dy}{dx} = -\frac{9x}{16y}. $$

在 $\big(2,\,\tfrac{3\sqrt{3}}{2}\big)$ 处斜率为 $-\dfrac{9(2)}{16\cdot\frac{3\sqrt 3}{2}} = -\dfrac{18}{24\sqrt 3} = -\dfrac{3}{4\sqrt 3} = -\dfrac{\sqrt 3}{4}$。因此切线为 $y - \tfrac{3\sqrt 3}{2} = -\tfrac{\sqrt 3}{4}\,(x - 2)$。若显式解出 $y$，就会引入一个平方根并需要选择分支，所以隐函数求导是更简洁的途径。

For the folium $x^3 + y^3 = 6xy$ of Example 1.2 we found $\dfrac{dy}{dx} = \dfrac{2y - x^2}{y^2 - 2x}$. Where is the tangent horizontal, and where is it vertical?

A horizontal tangent needs the numerator zero with the denominator nonzero: $2y - x^2 = 0$, so $y = \tfrac{x^2}{2}$. Substituting into the curve $x^3 + y^3 = 6xy$ gives $x^3 + \tfrac{x^6}{8} = 3x^3$, that is $\tfrac{x^6}{8} = 2x^3$, so $x^3 = 16$ and $x = 16^{1/3} = 2^{4/3}$. Then $y = \tfrac{1}{2}x^2 = 2^{5/3}$.

A vertical tangent needs the denominator zero with the numerator nonzero: $y^2 - 2x = 0$, so $x = \tfrac{y^2}{2}$, which by the symmetry $x \leftrightarrow y$ of the curve gives $y = 2^{4/3}$, $x = 2^{5/3}$. The point $(0,0)$ makes both numerator and denominator vanish, signalling the self-intersection where the curve crosses itself and has two tangent lines rather than one.

对例题 1.2 中的叶形线 $x^3 + y^3 = 6xy$，我们已求得 $\dfrac{dy}{dx} = \dfrac{2y - x^2}{y^2 - 2x}$。切线在哪里水平，又在哪里竖直？

水平切线要求分子为零而分母非零：$2y - x^2 = 0$，即 $y = \tfrac{x^2}{2}$。代入曲线 $x^3 + y^3 = 6xy$ 得 $x^3 + \tfrac{x^6}{8} = 3x^3$，即 $\tfrac{x^6}{8} = 2x^3$，于是 $x^3 = 16$，$x = 16^{1/3} = 2^{4/3}$。再由 $y = \tfrac{1}{2}x^2 = 2^{5/3}$。

竖直切线要求分母为零而分子非零：$y^2 - 2x = 0$，即 $x = \tfrac{y^2}{2}$；由曲线关于 $x \leftrightarrow y$ 的对称性，得 $y = 2^{4/3}$，$x = 2^{5/3}$。点 $(0,0)$ 使分子分母同时为零，标志着曲线的自交点——曲线在此与自身相交，拥有两条切线而非一条。

The legitimacy of treating $y$ as a differentiable function of $x$ is guaranteed by the Implicit Function Theorem. If $F(x,y)$ is continuously differentiable near a point $(a,b)$ with $F(a,b)=0$ and $\frac{\partial F}{\partial y}(a,b) \neq 0$, then near $(a,b)$ the equation $F(x,y)=0$ defines a unique differentiable function $y = g(x)$ with $g(a)=b$, and

$$ \frac{dy}{dx} = -\frac{F_x}{F_y}. $$

For the circle $F = x^2+y^2-25$ we have $F_y = 2y$, which vanishes at $(\pm 5, 0)$. Those are exactly the points where the tangent is vertical and $\frac{dy}{dx} = -x/y$ is undefined, consistent with the formula breaking down where $F_y = 0$.

Here is the one-line derivation of the quotient formula. Along the curve $F(x,g(x)) = 0$ holds identically in $x$, so differentiating with the multivariable chain rule gives $F_x + F_y\,g'(x) = 0$. Solving for $g'(x)$ recovers $\frac{dy}{dx} = -F_x/F_y$ whenever $F_y \neq 0$. This is exactly the bookkeeping of implicit differentiation, now justified rather than merely performed: the requirement $F_y \neq 0$ is precisely what licenses the division, and it fails at vertical tangents.

把 $y$ 当作 $x$ 的可微函数处理，其合法性由隐函数定理（Implicit Function Theorem）保证。若 $F(x,y)$ 在点 $(a,b)$ 附近连续可微，且 $F(a,b)=0$、$\frac{\partial F}{\partial y}(a,b) \neq 0$，则在 $(a,b)$ 附近方程 $F(x,y)=0$ 定义出唯一的可微函数 $y = g(x)$，满足 $g(a)=b$，且

$$ \frac{dy}{dx} = -\frac{F_x}{F_y}. $$

对圆 $F = x^2+y^2-25$，有 $F_y = 2y$，它在 $(\pm 5, 0)$ 处为零。这恰好是切线竖直、$\frac{dy}{dx} = -x/y$ 无定义的那些点，与公式在 $F_y = 0$ 处失效相一致。

下面是商式公式的一行推导。沿曲线，$F(x,g(x)) = 0$ 关于 $x$ 恒成立，所以用多元链式法则求导得 $F_x + F_y\,g'(x) = 0$。解出 $g'(x)$，只要 $F_y \neq 0$ 就回到 $\frac{dy}{dx} = -F_x/F_y$。这正是隐函数求导的记账过程，如今得到了证明而非仅仅被执行：条件 $F_y \neq 0$ 恰是允许做除法的依据，而它在竖直切线处失效。

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

$$ \cos y \,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\cos y}. $$

On the chosen range $\cos y \ge 0$, and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. Therefore

$$ \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1. $$

设 $y = \arcsin x$，则 $\sin y = x$，且 $y \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$。隐式求导：

$$ \cos y \,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\cos y}. $$

在所取的区间上 $\cos y \ge 0$，且 $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$。因此

$$ \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1. $$

For $y = \arctan x$ we have $\tan y = x$, so $\sec^2 y\,\frac{dy}{dx} = 1$. Since $\sec^2 y = 1 + \tan^2 y = 1 + x^2$,

$$ \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}. $$

For $y = \ln x$ we have $e^{y} = x$, so $e^{y}\,\frac{dy}{dx} = 1$, giving $\frac{dy}{dx} = e^{-y} = \frac{1}{x}$. Implicit differentiation thus recovers $\frac{d}{dx}\ln x = \frac{1}{x}$ from the exponential.

对 $y = \arctan x$，有 $\tan y = x$，故 $\sec^2 y\,\frac{dy}{dx} = 1$。又因 $\sec^2 y = 1 + \tan^2 y = 1 + x^2$，所以

$$ \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}. $$

对 $y = \ln x$，有 $e^{y} = x$，故 $e^{y}\,\frac{dy}{dx} = 1$，得 $\frac{dy}{dx} = e^{-y} = \frac{1}{x}$。可见隐函数求导从指数函数出发就重新得到了 $\frac{d}{dx}\ln x = \frac{1}{x}$。

Let $y = \operatorname{arcsec} x$ for $|x| > 1$, so $\sec y = x$ with $y \in [0,\tfrac{\pi}{2}) \cup (\tfrac{\pi}{2}, \pi]$. Differentiate implicitly:

$$ \sec y \tan y\,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\sec y \tan y}. $$

Now $\sec y = x$ and $\tan y = \pm\sqrt{\sec^2 y - 1} = \pm\sqrt{x^2 - 1}$. On the chosen branch $\sec y \tan y > 0$ for both $x>1$ and $x<-1$, so we keep the positive magnitude and write

$$ \frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2 - 1}}, \qquad |x| > 1. $$

The absolute value is the whole subtlety here: it is forced by the branch and is what makes the derivative positive on both pieces of the domain, matching the fact that $\operatorname{arcsec}$ is increasing on each piece.

设 $y = \operatorname{arcsec} x$，其中 $|x| > 1$，则 $\sec y = x$，且 $y \in [0,\tfrac{\pi}{2}) \cup (\tfrac{\pi}{2}, \pi]$。隐式求导：

$$ \sec y \tan y\,\frac{dy}{dx} = 1 \;\Longrightarrow\; \frac{dy}{dx} = \frac{1}{\sec y \tan y}. $$

此时 $\sec y = x$，且 $\tan y = \pm\sqrt{\sec^2 y - 1} = \pm\sqrt{x^2 - 1}$。在所取的分支上，无论 $x>1$ 还是 $x<-1$ 都有 $\sec y \tan y > 0$，所以我们保留正的量值，写成

$$ \frac{d}{dx}\operatorname{arcsec} x = \frac{1}{|x|\sqrt{x^2 - 1}}, \qquad |x| > 1. $$

绝对值正是这里全部的微妙之处：它由分支所迫，使导数在定义域的两段上都为正，这与 $\operatorname{arcsec}$ 在每一段上都单调递增的事实相吻合。

Let $f(x) = x^5 + 2x + 1$. This is strictly increasing (its derivative $5x^4 + 2$ is always positive), so it has an inverse $g = f^{-1}$, yet there is no elementary formula for $g$. Find $g'(4)$.

We need the point $a$ with $f(a) = 4$. Trying $a = 1$ gives $f(1) = 1 + 2 + 1 = 4$, so $a = 1$ and $g(4) = 1$. Then

$$ g'(4) = \frac{1}{f'(1)} = \frac{1}{5(1)^4 + 2} = \frac{1}{7}. $$

The inverse-function rule delivers the slope of a curve we cannot even write down explicitly, which is the practical payoff of the formula.

设 $f(x) = x^5 + 2x + 1$。它严格递增（其导数 $5x^4 + 2$ 恒为正），所以存在反函数 $g = f^{-1}$，但 $g$ 没有初等公式。求 $g'(4)$。

我们需要满足 $f(a) = 4$ 的点 $a$。试 $a = 1$ 得 $f(1) = 1 + 2 + 1 = 4$，于是 $a = 1$，$g(4) = 1$。则

$$ g'(4) = \frac{1}{f'(1)} = \frac{1}{5(1)^4 + 2} = \frac{1}{7}. $$

反函数求导法则给出了一条我们甚至无法显式写出的曲线的斜率，这正是该公式的实用价值。

Suppose $f$ is differentiable and one-to-one on an open interval, $g = f^{-1}$ is continuous at $b$, and $f'(a) \neq 0$ where $a = g(b)$. We prove $g'(b) = 1/f'(a)$ directly from the difference quotient. For $y \neq b$ near $b$, set $x = g(y)$ and $a = g(b)$; because $g$ is one-to-one, $x \neq a$, and

$$ \frac{g(y) - g(b)}{y - b} = \frac{x - a}{f(x) - f(a)} = \frac{1}{\dfrac{f(x) - f(a)}{x - a}}. $$

As $y \to b$, continuity of $g$ forces $x = g(y) \to g(b) = a$. The inner difference quotient tends to $f'(a) \neq 0$, so the reciprocal tends to $1/f'(a)$. Therefore the limit exists and

$$ g'(b) = \frac{1}{f'(a)} = \frac{1}{f'(g(b))}. $$

The hypothesis $f'(a)\neq 0$ is essential: where $f$ has a horizontal tangent, the inverse has a vertical tangent and $g'$ does not exist, exactly as the reciprocal would predict.

设 $f$ 在某开区间上可微且一一对应，$g = f^{-1}$ 在 $b$ 处连续（continuous），且在 $a = g(b)$ 处 $f'(a) \neq 0$。我们直接从差商出发证明 $g'(b) = 1/f'(a)$。对于 $b$ 附近 $y \neq b$，令 $x = g(y)$、$a = g(b)$；因 $g$ 一一对应，故 $x \neq a$，且

$$ \frac{g(y) - g(b)}{y - b} = \frac{x - a}{f(x) - f(a)} = \frac{1}{\dfrac{f(x) - f(a)}{x - a}}. $$

当 $y \to b$ 时，$g$ 的连续性迫使 $x = g(y) \to g(b) = a$。内层差商趋于 $f'(a) \neq 0$，所以其倒数趋于 $1/f'(a)$。因此极限（limit）存在，且

$$ g'(b) = \frac{1}{f'(a)} = \frac{1}{f'(g(b))}. $$

假设 $f'(a)\neq 0$ 至关重要：在 $f$ 有水平切线之处，反函数有竖直切线，$g'$ 不存在，这恰如倒数所预言的那样。

Air is pumped into a spherical balloon at $100\ \text{cm}^3/\text{s}$. How fast is the radius increasing when $r = 5\ \text{cm}$?

Relation: $V = \tfrac{4}{3}\pi r^3$. Differentiate in $t$:

$$ \frac{dV}{dt} = 4\pi r^2 \,\frac{dr}{dt}. $$

Substitute $\frac{dV}{dt} = 100$ and $r = 5$:

$$ 100 = 4\pi (25)\,\frac{dr}{dt} \;\Longrightarrow\; \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}. $$

以 $100\ \text{cm}^3/\text{s}$ 的速率向球形气球充气。当 $r = 5\ \text{cm}$ 时，半径增大得有多快？

关系式：$V = \tfrac{4}{3}\pi r^3$。对 $t$ 求导：

$$ \frac{dV}{dt} = 4\pi r^2 \,\frac{dr}{dt}. $$

代入 $\frac{dV}{dt} = 100$ 与 $r = 5$：

$$ 100 = 4\pi (25)\,\frac{dr}{dt} \;\Longrightarrow\; \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}. $$

With air still entering at $100\ \text{cm}^3/\text{s}$, how fast is the surface area $A = 4\pi r^2$ growing when $r = 5\ \text{cm}$?

We already know $\frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}$ at this instant from Example 3.1. Differentiate the area relation in $t$:

$$ \frac{dA}{dt} = 8\pi r\,\frac{dr}{dt}. $$

Substitute $r = 5$ and $\frac{dr}{dt} = \frac{1}{\pi}$:

$$ \frac{dA}{dt} = 8\pi (5)\Big(\frac{1}{\pi}\Big) = 40\ \text{cm}^2/\text{s}. $$

There is also a slick check: since $V = \tfrac{4}{3}\pi r^3$ gives $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ and $A = 4\pi r^2$, we have $\frac{dV}{dt} = A\frac{dr}{dt}$, hence $\frac{dr}{dt} = \frac{1}{A}\frac{dV}{dt}$. Then $\frac{dA}{dt} = 8\pi r\frac{dr}{dt} = \frac{2}{r}\frac{dV}{dt} = \frac{2}{5}(100) = 40\ \text{cm}^2/\text{s}$, matching exactly.

空气仍以 $100\ \text{cm}^3/\text{s}$ 进入，当 $r = 5\ \text{cm}$ 时，表面积 $A = 4\pi r^2$ 增长得有多快？

由例题 3.1，我们已知此刻 $\frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}$。对面积关系式关于 $t$ 求导：

$$ \frac{dA}{dt} = 8\pi r\,\frac{dr}{dt}. $$

代入 $r = 5$ 与 $\frac{dr}{dt} = \frac{1}{\pi}$：

$$ \frac{dA}{dt} = 8\pi (5)\Big(\frac{1}{\pi}\Big) = 40\ \text{cm}^2/\text{s}. $$

还有一个漂亮的验算：由 $V = \tfrac{4}{3}\pi r^3$ 得 $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$，又 $A = 4\pi r^2$，故 $\frac{dV}{dt} = A\frac{dr}{dt}$，于是 $\frac{dr}{dt} = \frac{1}{A}\frac{dV}{dt}$。则 $\frac{dA}{dt} = 8\pi r\frac{dr}{dt} = \frac{2}{r}\frac{dV}{dt} = \frac{2}{5}(100) = 40\ \text{cm}^2/\text{s}$，完全吻合。

A circular oil slick spreads so that its area grows at a constant $6\ \text{m}^2/\text{s}$. How fast is the radius growing when $r = 3\ \text{m}$, and again when $r = 6\ \text{m}$?

From $A = \pi r^2$, differentiate: $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$, so $\frac{dr}{dt} = \dfrac{1}{2\pi r}\frac{dA}{dt}$. With $\frac{dA}{dt} = 6$:

$$ \left.\frac{dr}{dt}\right|_{r=3} = \frac{6}{2\pi(3)} = \frac{1}{\pi} \approx 0.318\ \text{m/s}, \qquad \left.\frac{dr}{dt}\right|_{r=6} = \frac{6}{2\pi(6)} = \frac{1}{2\pi} \approx 0.159\ \text{m/s}. $$

Even though the area grows at a fixed rate, the radius grows more slowly as the slick widens, because the same new area is spread around an ever longer circumference. This is the qualitative lesson behind every related-rates problem with a nonlinear relation: equal rates in one quantity need not mean equal rates in another.

一片圆形油膜扩散，使其面积以恒定的 $6\ \text{m}^2/\text{s}$ 增长。当 $r = 3\ \text{m}$ 以及 $r = 6\ \text{m}$ 时，半径增长得有多快？

由 $A = \pi r^2$ 求导：$\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$，故 $\frac{dr}{dt} = \dfrac{1}{2\pi r}\frac{dA}{dt}$。代入 $\frac{dA}{dt} = 6$：

$$ \left.\frac{dr}{dt}\right|_{r=3} = \frac{6}{2\pi(3)} = \frac{1}{\pi} \approx 0.318\ \text{m/s}, \qquad \left.\frac{dr}{dt}\right|_{r=6} = \frac{6}{2\pi(6)} = \frac{1}{2\pi} \approx 0.159\ \text{m/s}. $$

尽管面积以固定速率增长，但随着油膜变宽，半径增长得越来越慢，因为同样大小的新增面积被铺展在越来越长的周长上。这正是每一个带非线性关系的相关变化率问题背后的定性启示：某一个量的变化率相等，并不意味着另一个量的变化率也相等。

A conical tank, point down, has height $12\ \text{m}$ and top radius $4\ \text{m}$. Water drains at $2\ \text{m}^3/\text{min}$. How fast is the depth falling when the water is $6\ \text{m}$ deep?

By similar triangles the surface radius $r$ and depth $h$ satisfy $\frac{r}{h} = \frac{4}{12} = \frac{1}{3}$, so $r = \frac{h}{3}$. The volume is

$$ V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi \Big(\tfrac{h}{3}\Big)^2 h = \frac{\pi h^3}{27}. $$

Differentiate in $t$: $\frac{dV}{dt} = \frac{\pi h^2}{9}\,\frac{dh}{dt}$. With $\frac{dV}{dt} = -2$ and $h = 6$:

$$ -2 = \frac{\pi (36)}{9}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = -\frac{1}{2\pi} \approx -0.159\ \text{m/min}. $$

The negative sign confirms the depth is decreasing.

一个尖端朝下的圆锥形水箱，高 $12\ \text{m}$，顶部半径 $4\ \text{m}$。水以 $2\ \text{m}^3/\text{min}$ 排出。当水深 $6\ \text{m}$ 时，水深下降得有多快？

由相似三角形，水面半径 $r$ 与水深 $h$ 满足 $\frac{r}{h} = \frac{4}{12} = \frac{1}{3}$，故 $r = \frac{h}{3}$。体积为

$$ V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi \Big(\tfrac{h}{3}\Big)^2 h = \frac{\pi h^3}{27}. $$

对 $t$ 求导：$\frac{dV}{dt} = \frac{\pi h^2}{9}\,\frac{dh}{dt}$。代入 $\frac{dV}{dt} = -2$ 与 $h = 6$：

$$ -2 = \frac{\pi (36)}{9}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = -\frac{1}{2\pi} \approx -0.159\ \text{m/min}. $$

负号确认水深正在减小。

A $13\ \text{ft}$ ladder leans on a wall; its base slides away at $2\ \text{ft/s}$. How fast is the top sliding down when the base is $5\ \text{ft}$ from the wall?

From $x^2 + y^2 = 169$, at $x = 5$ we get $y = 12$. Differentiate: $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$. With $\frac{dx}{dt} = 2$,

$$ 5(2) + 12\,\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6}\ \text{ft/s}. $$

The top descends at $\frac{5}{6}\ \text{ft/s}$.

一架 $13\ \text{ft}$ 长的梯子靠在墙上，其底部以 $2\ \text{ft/s}$ 向外滑动。当底部离墙 $5\ \text{ft}$ 时，顶端下滑得有多快？

由 $x^2 + y^2 = 169$，当 $x = 5$ 时得 $y = 12$。求导：$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$。代入 $\frac{dx}{dt} = 2$，

$$ 5(2) + 12\,\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6}\ \text{ft/s}. $$

顶端以 $\frac{5}{6}\ \text{ft/s}$ 下降。

A street lamp sits atop a $5\ \text{m}$ pole. A person $1.8\ \text{m}$ tall walks away from the pole at $1.5\ \text{m/s}$. How fast is the tip of the shadow moving when the person is $8\ \text{m}$ from the pole?

Let $x$ be the person distance from the pole and $s$ the length of the shadow beyond the person, so the shadow tip is at distance $x + s$ from the pole. Similar triangles, the big triangle from lamp to shadow tip and the small one from head to shadow tip, give

$$ \frac{5}{x + s} = \frac{1.8}{s} \;\Longrightarrow\; 5s = 1.8(x+s) \;\Longrightarrow\; 3.2\,s = 1.8\,x \;\Longrightarrow\; s = \frac{9}{16}x. $$

The tip position is $p = x + s = x + \tfrac{9}{16}x = \tfrac{25}{16}x$. Differentiate in $t$:

$$ \frac{dp}{dt} = \frac{25}{16}\,\frac{dx}{dt} = \frac{25}{16}(1.5) = \frac{75}{32} \approx 2.34\ \text{m/s}. $$

Notice the answer does not depend on the value $x = 8$: the tip moves at a constant speed because the relation between $p$ and $x$ is linear. The person distance was a distractor, a deliberately common feature of exam problems.

一盏路灯位于 $5\ \text{m}$ 高的灯柱顶端。一个身高 $1.8\ \text{m}$ 的人以 $1.5\ \text{m/s}$ 远离灯柱行走。当此人离灯柱 $8\ \text{m}$ 时，影子的尖端移动得有多快？

设 $x$ 为此人离灯柱的距离，$s$ 为此人身后影子的长度，于是影子尖端离灯柱的距离为 $x + s$。由相似三角形——从灯到影子尖端的大三角形与从头顶到影子尖端的小三角形——得

$$ \frac{5}{x + s} = \frac{1.8}{s} \;\Longrightarrow\; 5s = 1.8(x+s) \;\Longrightarrow\; 3.2\,s = 1.8\,x \;\Longrightarrow\; s = \frac{9}{16}x. $$

尖端位置为 $p = x + s = x + \tfrac{9}{16}x = \tfrac{25}{16}x$。对 $t$ 求导：

$$ \frac{dp}{dt} = \frac{25}{16}\,\frac{dx}{dt} = \frac{25}{16}(1.5) = \frac{75}{32} \approx 2.34\ \text{m/s}. $$

注意答案并不依赖于 $x = 8$ 这个数值：尖端以恒定速率移动，因为 $p$ 与 $x$ 之间的关系是线性的。此人离灯柱的距离是一个干扰项，这是考题刻意设置的常见特征。

An observer stands $100\ \text{m}$ from the launch point of a balloon that rises straight up at $4\ \text{m/s}$. How fast is the angle of elevation $\theta$ increasing when the balloon is $100\ \text{m}$ high?

Let $h$ be the height. Then $\tan\theta = \dfrac{h}{100}$. Differentiate in $t$, remembering the chain rule on $\tan\theta$:

$$ \sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{100}\,\frac{dh}{dt}. $$

When $h = 100$, the triangle is isosceles, $\theta = \tfrac{\pi}{4}$, and $\sec^2\theta = 2$. With $\frac{dh}{dt} = 4$:

$$ 2\,\frac{d\theta}{dt} = \frac{4}{100} \;\Longrightarrow\; \frac{d\theta}{dt} = \frac{1}{50} = 0.02\ \text{rad/s}. $$

一名观察者站在距气球发射点 $100\ \text{m}$ 处，气球以 $4\ \text{m/s}$ 竖直上升。当气球高 $100\ \text{m}$ 时，仰角 $\theta$ 增大得有多快？

设 $h$ 为高度。则 $\tan\theta = \dfrac{h}{100}$。对 $t$ 求导，记得对 $\tan\theta$ 用链式法则：

$$ \sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{100}\,\frac{dh}{dt}. $$

当 $h = 100$ 时，三角形为等腰三角形，$\theta = \tfrac{\pi}{4}$，$\sec^2\theta = 2$。代入 $\frac{dh}{dt} = 4$：

$$ 2\,\frac{d\theta}{dt} = \frac{4}{100} \;\Longrightarrow\; \frac{d\theta}{dt} = \frac{1}{50} = 0.02\ \text{rad/s}. $$

Car A is $0.3\ \text{km}$ west of an intersection moving east at $60\ \text{km/h}$; car B is $0.4\ \text{km}$ south moving north at $80\ \text{km/h}$. How fast is the distance between them changing at this instant?

Let $x$ and $y$ be the distances of A and B from the intersection. Since both close in, $\frac{dx}{dt} = -60$ and $\frac{dy}{dt} = -80$. With $x = 0.3$, $y = 0.4$, the separation is $s = \sqrt{0.3^2 + 0.4^2} = 0.5$. From $s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$:

$$ 0.5\,\frac{ds}{dt} = (0.3)(-60) + (0.4)(-80) = -18 - 32 = -50, $$ $$ \frac{ds}{dt} = -100\ \text{km/h}. $$

The cars are approaching at $100\ \text{km/h}$.

A 车在交叉路口以西 $0.3\ \text{km}$ 处，以 $60\ \text{km/h}$ 向东行驶；B 车在以南 $0.4\ \text{km}$ 处，以 $80\ \text{km/h}$ 向北行驶。此刻两车之间的距离变化得有多快？

设 $x$、$y$ 分别为 A、B 离交叉路口的距离。由于两车都在靠近，$\frac{dx}{dt} = -60$，$\frac{dy}{dt} = -80$。当 $x = 0.3$、$y = 0.4$ 时，间距为 $s = \sqrt{0.3^2 + 0.4^2} = 0.5$。由 $s\frac{ds}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$：

$$ 0.5\,\frac{ds}{dt} = (0.3)(-60) + (0.4)(-80) = -18 - 32 = -50, $$ $$ \frac{ds}{dt} = -100\ \text{km/h}. $$

两车正以 $100\ \text{km/h}$ 相互靠近。

A tank holds $100\ \text{L}$ of pure water. Brine containing $0.2\ \text{kg/L}$ of salt flows in at $5\ \text{L/min}$, and the well stirred mixture flows out at the same rate. Let $S(t)$ be the kilograms of salt present. Find $\frac{dS}{dt}$ when $S = 4\ \text{kg}$.

Conservation of mass gives rate in minus rate out. The inflow carries $0.2 \times 5 = 1\ \text{kg/min}$. The concentration in the tank is $\frac{S}{100}\ \text{kg/L}$, so the outflow carries $\frac{S}{100}\times 5 = \frac{S}{20}\ \text{kg/min}$:

$$ \frac{dS}{dt} = 1 - \frac{S}{20}. $$

At $S = 4$: $\frac{dS}{dt} = 1 - \frac{4}{20} = 1 - 0.2 = 0.8\ \text{kg/min}$.

一个槽里装有 $100\ \text{L}$ 纯水。含盐 $0.2\ \text{kg/L}$ 的盐水以 $5\ \text{L/min}$ 流入，充分搅拌后的混合液以相同速率流出。设 $S(t)$ 为槽中盐的千克数。求 $S = 4\ \text{kg}$ 时的 $\frac{dS}{dt}$。

质量守恒给出流入速率减去流出速率。流入携带 $0.2 \times 5 = 1\ \text{kg/min}$。槽内浓度为 $\frac{S}{100}\ \text{kg/L}$，所以流出携带 $\frac{S}{100}\times 5 = \frac{S}{20}\ \text{kg/min}$：

$$ \frac{dS}{dt} = 1 - \frac{S}{20}. $$

当 $S = 4$ 时：$\frac{dS}{dt} = 1 - \frac{4}{20} = 1 - 0.2 = 0.8\ \text{kg/min}$。

A police car sits $40\ \text{m}$ off a straight highway. A car drives along the highway, and when it is $30\ \text{m}$ down the road from the nearest point, the line-of-sight distance from cruiser to car is $50\ \text{m}$ and is increasing at $20\ \text{m/s}$. How fast is the car actually travelling along the road?

Let $x$ be the car distance along the road from the nearest point and $D$ the line-of-sight distance. Then $D^2 = 40^2 + x^2$. Differentiate in $t$:

$$ 2D\,\frac{dD}{dt} = 2x\,\frac{dx}{dt} \;\Longrightarrow\; D\,\frac{dD}{dt} = x\,\frac{dx}{dt}. $$

At the instant given $x = 30$ (since $30^2 + 40^2 = 50^2$), $D = 50$, $\frac{dD}{dt} = 20$:

$$ 50(20) = 30\,\frac{dx}{dt} \;\Longrightarrow\; \frac{dx}{dt} = \frac{1000}{30} \approx 33.3\ \text{m/s}. $$

The car true speed exceeds the $20\ \text{m/s}$ closing rate the radar reads, because the radar measures only the component of velocity along the line of sight; the geometry recovers the full road speed.

一辆警车停在距直线公路 $40\ \text{m}$ 处。一辆车沿公路行驶，当它在距最近点 $30\ \text{m}$ 处时，警车到该车的视线距离为 $50\ \text{m}$，并以 $20\ \text{m/s}$ 增大。该车沿公路的真实行驶速率是多少？

设 $x$ 为该车沿公路距最近点的距离，$D$ 为视线距离。则 $D^2 = 40^2 + x^2$。对 $t$ 求导：

$$ 2D\,\frac{dD}{dt} = 2x\,\frac{dx}{dt} \;\Longrightarrow\; D\,\frac{dD}{dt} = x\,\frac{dx}{dt}. $$

在给定瞬间 $x = 30$（因为 $30^2 + 40^2 = 50^2$），$D = 50$，$\frac{dD}{dt} = 20$：

$$ 50(20) = 30\,\frac{dx}{dt} \;\Longrightarrow\; \frac{dx}{dt} = \frac{1000}{30} \approx 33.3\ \text{m/s}. $$

该车的真实速率超过雷达读到的 $20\ \text{m/s}$ 接近速率，因为雷达只测量速度沿视线方向的分量；通过几何关系才恢复出完整的公路速率。

Now suppose the tank in Example 5.2 starts with $100\ \text{L}$ but the outflow is only $3\ \text{L/min}$ while inflow stays at $5\ \text{L/min}$, so the volume rises. Write the rate equation for the salt $S(t)$ at a general time $t$.

The volume after $t$ minutes is $V(t) = 100 + (5 - 3)t = 100 + 2t$ litres. Inflow still carries $0.2 \times 5 = 1\ \text{kg/min}$. The outflow concentration is $\dfrac{S}{V} = \dfrac{S}{100 + 2t}$, and it leaves at $3\ \text{L/min}$, so the outflow rate of salt is $\dfrac{3S}{100 + 2t}$:

$$ \frac{dS}{dt} = 1 - \frac{3S}{100 + 2t}. $$

This is now a genuine differential equation rather than a single-instant rate, because the denominator carries an explicit $t$. The key modelling step is that the outflow concentration uses the current volume $V(t)$, not the initial $100\ \text{L}$; freezing the volume is the most common error in variable-volume mixing.

现在假设例题 5.2 中的槽起初有 $100\ \text{L}$，但流出仅为 $3\ \text{L/min}$，而流入仍为 $5\ \text{L/min}$，因此体积上升。写出盐 $S(t)$ 在一般时刻 $t$ 的变化率方程。

经过 $t$ 分钟后体积为 $V(t) = 100 + (5 - 3)t = 100 + 2t$ 升。流入仍携带 $0.2 \times 5 = 1\ \text{kg/min}$。流出浓度为 $\dfrac{S}{V} = \dfrac{S}{100 + 2t}$，以 $3\ \text{L/min}$ 流出，所以盐的流出速率为 $\dfrac{3S}{100 + 2t}$：

$$ \frac{dS}{dt} = 1 - \frac{3S}{100 + 2t}. $$

这现在是一个真正的微分方程（differential equation），而非单一瞬间的变化率，因为分母带有显式的 $t$。关键的建模步骤在于：流出浓度使用当前体积 $V(t)$，而不是初始的 $100\ \text{L}$；冻结体积是变体积混合问题中最常见的错误。

Differentiate $x\,e^{y} = x + y$ implicitly.

The left side needs the product rule, and $e^{y}$ needs the chain rule:

$$ e^{y} + x\,e^{y}\,\frac{dy}{dx} = 1 + \frac{dy}{dx}. $$

Collect: $\big(x e^{y} - 1\big)\frac{dy}{dx} = 1 - e^{y}$, so $\frac{dy}{dx} = \frac{1 - e^{y}}{x e^{y} - 1}$. Forgetting the chain rule on $e^{y}$ or the product rule on $x e^{y}$ both give a wrong answer.

对 $x\,e^{y} = x + y$ 隐式求导。

左边需要乘积法则，而 $e^{y}$ 需要链式法则：

$$ e^{y} + x\,e^{y}\,\frac{dy}{dx} = 1 + \frac{dy}{dx}. $$

归并：$\big(x e^{y} - 1\big)\frac{dy}{dx} = 1 - e^{y}$，故 $\frac{dy}{dx} = \frac{1 - e^{y}}{x e^{y} - 1}$。在 $e^{y}$ 上漏掉链式法则，或在 $x e^{y}$ 上漏掉乘积法则，都会得到错误答案。

A worked answer should pass a units check before you trust it. Return to the draining cone (Example 4.1), where $\frac{dh}{dt} = -\frac{1}{2\pi}\ \text{m/min}$. Verify the units of the differentiated relation $\frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$.

The left side $\frac{dV}{dt}$ has units $\text{m}^3/\text{min}$. On the right, $\frac{\pi h^2}{9}$ has units $\text{m}^2$ (the $9$ and $\pi$ are pure numbers, since the original $\frac{r}{h} = \frac13$ ratio is dimensionless), and $\frac{dh}{dt}$ has units $\text{m}/\text{min}$. The product is $\text{m}^2 \cdot \text{m}/\text{min} = \text{m}^3/\text{min}$, matching the left side. A mismatch would flag an algebra slip immediately, well before you reach a numerical answer.

一个算出来的答案在你信任它之前，应当先通过单位检验。回到排水圆锥（例题 4.1），其中 $\frac{dh}{dt} = -\frac{1}{2\pi}\ \text{m/min}$。验证求导后关系式 $\frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt}$ 的单位。

左边 $\frac{dV}{dt}$ 的单位是 $\text{m}^3/\text{min}$。右边，$\frac{\pi h^2}{9}$ 的单位是 $\text{m}^2$（$9$ 与 $\pi$ 都是纯数，因为原比例 $\frac{r}{h} = \frac13$ 是无量纲的），而 $\frac{dh}{dt}$ 的单位是 $\text{m}/\text{min}$。两者乘积为 $\text{m}^2 \cdot \text{m}/\text{min} = \text{m}^3/\text{min}$，与左边吻合。一旦不匹配，就会在你算出数值答案之前立即提示某处代数出了错。

In the sliding ladder (Example 4.2) the base moves away at $\frac{dx}{dt} = +2\ \text{ft/s}$ and we found $\frac{dy}{dt} = -\frac{5}{6}\ \text{ft/s}$. A student who writes $\frac{dx}{dt} = -2$ by mistake would get $\frac{dy}{dt} = +\frac{5}{6}$, claiming the top of the ladder rises while the base slides out, which is physically impossible.

The remedy is a sign convention fixed at the start: choose increasing $x$ to mean the base moving away from the wall, so $\frac{dx}{dt} > 0$; then the constraint $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$ forces $\frac{dy}{dt} < 0$, the top falling. Whenever a related-rates answer has a sign that contradicts the physical setup, the error is almost always a mis-signed given rate.

在下滑的梯子（例题 4.2）中，底部以 $\frac{dx}{dt} = +2\ \text{ft/s}$ 向外移动，我们求得 $\frac{dy}{dt} = -\frac{5}{6}\ \text{ft/s}$。若某学生误写 $\frac{dx}{dt} = -2$，就会得到 $\frac{dy}{dt} = +\frac{5}{6}$，声称底部向外滑动的同时梯顶却在上升，这在物理上不可能。

补救办法是一开始就固定符号约定：取 $x$ 增大表示底部远离墙，于是 $\frac{dx}{dt} > 0$；则约束 $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$ 迫使 $\frac{dy}{dt} < 0$，即梯顶下落。每当一个相关变化率答案的符号与物理情形矛盾时，错误几乎总是某个给定变化率的符号弄反了。

For $x^2 + y^2 = 1$ we found $\frac{dy}{dx} = -\frac{x}{y}$. Differentiate again with the quotient rule, remembering $\frac{dy}{dx}$ inside:

$$ \frac{d^2 y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}. $$

Since $x^2 + y^2 = 1$ on the curve, this simplifies to $\frac{d^2 y}{dx^2} = -\frac{1}{y^3}$. The sign matches the curvature of the circle: concave down on the upper half ($y>0$), concave up on the lower half.

对 $x^2 + y^2 = 1$，我们求得 $\frac{dy}{dx} = -\frac{x}{y}$。再用商的法则（Quotient Rule）求一次导，记得里面有 $\frac{dy}{dx}$：

$$ \frac{d^2 y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}. $$

因为在曲线上 $x^2 + y^2 = 1$，这化简为 $\frac{d^2 y}{dx^2} = -\frac{1}{y^3}$。其符号与圆的凹凸性（concavity）相符：上半部分（$y>0$）下凹，下半部分上凹。

Find $\frac{dy}{dx}$ for $y = x^{x}$ with $x>0$. Taking logarithms, $\ln y = x \ln x$. Differentiate implicitly:

$$ \frac{1}{y}\frac{dy}{dx} = \ln x + x\cdot\frac{1}{x} = \ln x + 1. $$

Therefore $\frac{dy}{dx} = y(\ln x + 1) = x^{x}(\ln x + 1)$. Neither the power rule nor the exponential rule applies directly here, because both the base and the exponent vary; logarithmic differentiation handles it cleanly.

对 $y = x^{x}$（$x>0$）求 $\frac{dy}{dx}$。两边取对数，$\ln y = x \ln x$。隐式求导：

$$ \frac{1}{y}\frac{dy}{dx} = \ln x + x\cdot\frac{1}{x} = \ln x + 1. $$

因此 $\frac{dy}{dx} = y(\ln x + 1) = x^{x}(\ln x + 1)$。这里幂函数法则和指数函数法则都不能直接套用，因为底数和指数都在变化；对数求导法干净利落地处理了它。

Differentiate $y = \dfrac{x^2\,\sqrt{x^2 + 1}}{(3x - 1)^4}$ using logarithmic differentiation.

Take logarithms and use $\ln$ to turn products, quotients, and powers into sums and multiples:

$$ \ln y = 2\ln x + \tfrac{1}{2}\ln(x^2 + 1) - 4\ln(3x - 1). $$

Differentiate each term, with the chain rule throughout:

$$ \frac{1}{y}\frac{dy}{dx} = \frac{2}{x} + \frac{x}{x^2 + 1} - \frac{12}{3x - 1}. $$

Multiply back by $y$:

$$ \frac{dy}{dx} = \frac{x^2\sqrt{x^2 + 1}}{(3x - 1)^4}\left( \frac{2}{x} + \frac{x}{x^2 + 1} - \frac{12}{3x - 1} \right). $$

Doing this by the product and quotient rules directly would mean differentiating a four-factor expression and simplifying a forest of terms; the logarithm flattens it into three independent pieces.

用对数求导法对 $y = \dfrac{x^2\,\sqrt{x^2 + 1}}{(3x - 1)^4}$ 求导。

取对数，用 $\ln$ 把乘积、商和幂化成和式与倍数：

$$ \ln y = 2\ln x + \tfrac{1}{2}\ln(x^2 + 1) - 4\ln(3x - 1). $$

逐项求导，全程使用链式法则：

$$ \frac{1}{y}\frac{dy}{dx} = \frac{2}{x} + \frac{x}{x^2 + 1} - \frac{12}{3x - 1}. $$

再乘回 $y$：

$$ \frac{dy}{dx} = \frac{x^2\sqrt{x^2 + 1}}{(3x - 1)^4}\left( \frac{2}{x} + \frac{x}{x^2 + 1} - \frac{12}{3x - 1} \right). $$

若直接用乘积法则和商的法则来做，就意味着对一个四因子表达式求导，并简化一大堆项；而对数把它压平成三个相互独立的部分。

For $x^3 + y^3 = 6xy$ find $\dfrac{d^2 y}{dx^2}$ at the point $(3,3)$, where we found $\dfrac{dy}{dx} = -1$ in Example 1.2.

Differentiate $3x^2 + 3y^2 y' = 6(y + x y')$ once more in $x$, treating $y'$ as a function of $x$ and using the product rule on $y^2 y'$ and on $x y'$:

$$ 6x + 6y\,(y')^2 + 3y^2\,y'' = 6\big(2y' + x\,y''\big). $$

At $(3,3)$ with $y' = -1$, substitute the numbers:

$$ 18 + 18(1) + 27\,y'' = 6\big(-2 + 3\,y''\big) = -12 + 18\,y''. $$

So $36 + 27 y'' = -12 + 18 y''$, giving $9 y'' = -48$ and $\dfrac{d^2 y}{dx^2} = -\dfrac{16}{3}$. The negative value says the curve is concave down at $(3,3)$, consistent with the loop of the folium bending away from the tangent line $y = -x + 6$.

对 $x^3 + y^3 = 6xy$，在点 $(3,3)$ 处求 $\dfrac{d^2 y}{dx^2}$；在例题 1.2 中我们已求得该处 $\dfrac{dy}{dx} = -1$。

对 $3x^2 + 3y^2 y' = 6(y + x y')$ 再关于 $x$ 求一次导，把 $y'$ 当作 $x$ 的函数，并对 $y^2 y'$ 和 $x y'$ 用乘积法则：

$$ 6x + 6y\,(y')^2 + 3y^2\,y'' = 6\big(2y' + x\,y''\big). $$

在 $(3,3)$ 处、$y' = -1$，代入数值：

$$ 18 + 18(1) + 27\,y'' = 6\big(-2 + 3\,y''\big) = -12 + 18\,y''. $$

于是 $36 + 27 y'' = -12 + 18 y''$，得 $9 y'' = -48$，$\dfrac{d^2 y}{dx^2} = -\dfrac{16}{3}$。负值表明曲线在 $(3,3)$ 处下凹，这与叶形线的环朝着远离切线 $y = -x + 6$ 的方向弯曲相一致。

For a positive differentiable function $y(x)$, the quantity $\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{d}{dx}\ln y$ is called the logarithmic derivative, and it measures the relative (fractional) rate of change of $y$. Its central property is that it converts multiplication into addition:

$$ \frac{(uv)'}{uv} = \frac{u'v + uv'}{uv} = \frac{u'}{u} + \frac{v'}{v}, \qquad \frac{(u/v)'}{u/v} = \frac{u'}{u} - \frac{v'}{v}. $$

These are exactly the product and quotient rules, rewritten in logarithmic form, and they are why $\ln$ of a product becomes a sum of logarithmic derivatives. For a power, $\dfrac{(u^n)'}{u^n} = n\dfrac{u'}{u}$, which is the chain rule on $u^n$. So every algebraic structure (product, quotient, power) collapses to addition under the logarithmic derivative, and the general formula for $y = f(x)^{g(x)}$ in the formula box above is just the product rule applied to $g(x)\ln f(x)$. This single identity is the engine behind both the $x^x$ computation and the product-quotient tower of Example 7.3.

对于一个正的可微函数 $y(x)$，量 $\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{d}{dx}\ln y$ 称为对数导数，它度量 $y$ 的相对（分数）变化率。其核心性质是把乘法转化为加法：

$$ \frac{(uv)'}{uv} = \frac{u'v + uv'}{uv} = \frac{u'}{u} + \frac{v'}{v}, \qquad \frac{(u/v)'}{u/v} = \frac{u'}{u} - \frac{v'}{v}. $$

这恰是用对数形式改写的乘积法则与商的法则，也正是为什么乘积的 $\ln$ 变成各个对数导数之和。对于幂，$\dfrac{(u^n)'}{u^n} = n\dfrac{u'}{u}$，这就是 $u^n$ 上的链式法则。于是在对数导数之下，每一种代数结构（乘积、商、幂）都坍缩为加法，而上面公式框中 $y = f(x)^{g(x)}$ 的一般公式，不过是对 $g(x)\ln f(x)$ 应用乘积法则。这一个恒等式正是 $x^x$ 计算和例题 7.3 中乘积-商多层结构背后的引擎。