3.6 Implicit Differentiation隐微分法，或者说隐函数求导法。对于一类函数（有时候也不是函数，如

隐微分法，或者说隐函数求导法。对于一类函数（有时候也不是函数，如果按照函数的定义）或方程，它不容易写成y = f(x)这样的形式，或者说这种形式不容易求导，那么怎么求导呢。

结合前面几节的各种求导方式，可以推导出对这类情形的处理方式，其中一个重要的思想是链式法则。如果将式子中的y看成f(x)，那么式子中就只剩下x，就可以应用求导法则了。

比如：

$y^2 = x$

假设y = f(x)，那么上式可以转化为：

$f(x)^2 = x$

将左右两边看作两个函数，既然它们在x上处处相等，它们的导数也相等，对两边分别对x求导：

\begin{aligned} \frac{d}{dx}[f(x)^2] &= \frac{d}{dx}(x) \\ 2f(x)\frac{dy}{dx} &= 1 \\ \frac{dy}{dx} &= \frac{1}{2y} \end{aligned}

注意：

$\frac{d}{dx}[f(x)^2] = 2f(x)\frac{dy}{dx}$

是应用了链式法则。

另一个更复杂的例子：

\begin{aligned} y^2 &= x^2 + \sin xy \\ \frac{d}{dx}y^2 &= \frac{d}{dx}x^2 + \frac{d}{dx}\sin xy \\ 2y\frac{dy}{dx} &= 2x + \cos xy\frac{d}{dx}(xy), \ 这里假设u = xy，应用链式法则，再对xy应用乘法法则 \\ 2y\frac{dy}{dx} &= 2x + \cos xy(y + x\frac{dy}{dx}) \\ \frac{dy}{dx} &= \frac{2x + y\cos xy}{2y - x\cos xy} \end{aligned}

高阶导数的例子，求 $2x^3 - 3y^2 = 8$ 对于x的二阶导数：

先求一阶导数：

\begin{aligned} \frac{d}{dx}(2x^3 - 3y^2) &= \frac{d}{dx}(8) \\ 6x^2 - 6y\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} = \frac{x^2}{y} \end{aligned}

再求二阶导数：

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(\frac{dy}{dx}) \\ &= \frac{d}{dx}(\frac{x^2}{y}) \\ &= \frac{2xy - x^2\frac{dy}{dx}}{y^2} \end{aligned}

代入 $dy/dx$ ，即得到最终结果：

$\frac{d^2y}{dx^2} = \frac{2x}{y} - \frac{x^4}{y^3}$

Power rule的拓展，前面的求导法则里，对于power rule要求指数是整数，但其实可以将其拓展到有理数。

证明：

设有整数p和q，函数 $y = x^{p/q} = \sqrt[q]{x^p}$ 可以转换为：

$y^q = x^p$

应用上面隐函数求导法则：

$qy^{q-1}\frac{dy}{dx} = px^{p-1}$

可得：

\begin{aligned} \frac{dy}{dx} &= \frac{px^{p-1}}{qy^{q-1}} \\ &= \frac{p}{q} \cdot \frac{x^{p-1}}{(x^{p/q})^{q-1}}, 将y代入 \\ &= \frac{p}{q} \cdot \frac{x^{p-1}}{x^{p - p/q}} \\ &= \frac{p}{q} \cdot x^{(p-1) - (p-p/q)} \\ &= \frac{p}{q} \cdot x^{(p/q) - 1} \end{aligned}

证毕

练习

$y = x^{9/4}$

$y' = \frac{9}{4}x^{5/4}$

$y = x^{-3/5}$

$y' = -\frac{3}{5}x^{-8/5}$

$y = \sqrt[3]{2x}$

$y' = \frac{2}{3}(2x)^{-2/3}$

$y = \sqrt[4]{5x}$

$y' = \frac{5}{4}(5x)^{-3/4}$

$y = 7\sqrt{x + 6}$

$y' = \frac{7}{2}(x + 6)^{-1/2}$

$y = -2\sqrt{x - 1}$

$y' = -(x - 1)^{-1/2}$

$y = (2x + 5)^{-1/2}$

$y' = -(2x + 5)^{-3/2}$

$y = (1 - 6x)^{2/3}$

$y' = -4(1 - 6x)^{-1/3}$

$y = x(x^2 + 1)^{1/2}$

$y' = (x^2 + 1)^{1/2} + x\frac{1}{2}(x^2 + 1)^{-1/2}2x = (2x^2 + 1)(x^2 + 1)^{-1/2}$

$y = x(x^2 + 1)^{-1/2}$

$y' = (x^2 + 1)^{-1/2} - x\frac{1}{2}(x^2 + 1)^{-3/2}2x = (x^2 + 1)^{-1/2} -x^2(x^2 + 1)^{-3/2}$

$s = \sqrt[7]{t^2}$

$s' = \frac{2}{7}t^{-5/7}$

$r = \sqrt[4]{\phi^{-3}}$

$r' = -\frac{3}{4}\phi^{-7/4}$

$y = \sin[(2t + 5)^{-2/3}]$

\begin{aligned} y' &= \cos[(2t + 5)^{-2/3}]\cdot \frac{d}{dt}(2t + 5)^{-2/3} \\ &= \cos[(2t + 5)^{-2/3}]\cdot -\frac{2}{3}(2t + 5)^{-5/3}\cdot \frac{d}{dt}(2t + 5) \\ &= \cos[(2t + 5)^{-2/3}]\cdot -\frac{2}{3}(2t + 5)^{-5/3}\cdot 2 \\ &= -\frac{4}{3}(2t + 5)^{-5/3}\cdot \cos[(2t + 5)^{-2/3}] \end{aligned}

$z = \cos[(1 - 6t)^{2/3}]$

$z' = -\sin[(1 - 6t)^{2/3}]\cdot -4(1 - 6t)^{-1/3} = 4(1 - 6t)^{-1/3}\cdot sin[(1 - 6t)^{2/3}]$

$f(x) = \sqrt{1 - \sqrt{x}}$

$f'(x) = \frac{1}{2}(1 - \sqrt{x})^{-1/2}\cdot -\frac{1}{2}x^{-1/2} = -\frac{1}{4}x^{-1/2}(1 - \sqrt{x})^{-1/2}$

$g(x) = 2(2x^{-1/2} + 1)^{-1/3}$

$g'(x) = -\frac{2}{3}(2x^{-1/2} + 1)\cdot -x^{-3/2} = \frac{4}{3}x^{-5/2} + \frac{2}{3}x^{-3/2}$

$h(\phi) = \sqrt[3]{1 + \cos 2\phi}$

$h'(\phi) = \frac{1}{3}(1 + \cos 2\phi)\cdot -2\sin 2\phi = -\frac{2}{3}\sin 2\phi (1 + \cos 2\phi)$

$k(\phi) = (\sin(\phi + 5))^{5/4}$

$k'(\phi) = \frac{5}{4}(\sin(\phi + 5))^{1/4}\cdot \cos(\phi + 5)$

$x^2y + xy^2 = 6$

\begin{aligned} \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) &= \frac{d}{dx}(6) \\ 2xy + 2x\frac{dy}{dx} + y^2 + 2x\frac{dy}{dx} &= 0 \\ 4x\frac{dy}{dx} &= -2xy - y^2 \\ \frac{dy}{dx} &= \frac{-2xy - y^2}{4x} \end{aligned}

$x^3 + y^3 = 18xy$

\begin{aligned} 3x^2 + 3y^2\frac{dy}{dx} &= 18y + 18x\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{6y - x^2}{y^2 - 6x} \end{aligned}

$2xy + y^2 = x + y$

\begin{aligned} 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} &= 1 + \frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{1 - 2y}{2x + 2y - 1} \end{aligned}

$x^3 - xy + y^3 = 1$

\begin{aligned} 2x^2 - y - x\frac{dy}{dx} + 3y^2\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= \frac{y - 2x^2}{3y^2 - x} \end{aligned}

$x^2(x - y)^2 = x^2 - y^2$

\begin{aligned} 2x(x - y)^2 + 2x^2\frac{d}{dx}(x - y) &= 2x - 2y\frac{dy}{dx} \\ 2x(x - y)^2 +2x^2(1 - \frac{dy}{dx}) &= 2x - 2y\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{x(x - y)^2 + x^2 - x}{x^2 - y} \end{aligned}

$(3xy + 7)^2 = 6y$

\begin{aligned} 2(3xy + 7)\frac{d}{dx}(3xy + 7) &= 6\frac{dy}{dx} \\ 2(3xy + 7)(3y + 3x\frac{dy}{dx}) &= 6\frac{dy}{dx} \\ 9xy^2 + 9x^2y\frac{dy}{dx} + 21y + 21x\frac{dy}{dx} &= 3\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{-3xy^2 - 7y}{3x^2y + 7x - 1} \end{aligned}

$y^2 = \frac{x - 1}{x + 1}$

\begin{aligned} 2y\frac{dy}{dx} &= \frac{(x + 1) - (x - 1)}{(x + 1)^2} \\ \frac{dy}{dx} &= \frac{1}{y(x + 1)^2} \end{aligned}

$x^2 = \frac{x - y}{x + y}$

\begin{aligned} 2x &= \frac{(1 - \frac{dy}{dx})(x + y) - (x - y)(1 + \frac{dy}{dx})}{(x + y)^2} \\ 2x(x + y)^2 &= x + y - x\frac{dy}{dx} - y\frac{dy}{dx} - x - x\frac{dy}{dx} + y + y\frac{dy}{dx} \\ 2x(x + y)^2 &= 2y - 2x\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{y - x(x + y)^2}{x} \end{aligned}

$x = \tan y$

\begin{aligned} 1 &= \sec^2 y\frac{dy}{dx} \\ \frac{dy}{dx} & = \frac{1}{\sec^2 y} \end{aligned}

$xy = \cot(xy)$

\begin{aligned} y + x\frac{dy}{dx} &= -\csc^2(xy)\cdot (y + x\frac{dy}{dx}) \\ \frac{dy}{dx} &= -\frac{y + y\csc^2(xy)}{x + x\csc^2(xy)} \end{aligned}

$x + \tan(xy) = 0$

\begin{aligned} 1 + \sec^2(xy)\cdot(y + x\frac{dy}{dx}) &= 0 \\ \frac{dy}{dx} &= -\frac{1 + y\sec^2(xy)}{x\sec^2(xy)} \end{aligned}

$x + \sin y = xy$

\begin{aligned} 1 + \cos y \cdot \frac{dy}{dx} &= y + x\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{y - 1}{\cos y - x} \end{aligned}

$y\sin(1/y) = 1 - xy$

\begin{aligned} \frac{dy}{dx}\sin(1/y) + y\cos(1/y)\cdot \frac{-\frac{dy}{dx}}{y^2} &= 0 - y - x\frac{dy}{dx} \\ \frac{dy}{dx} &= -\frac{y}{\sin(1/y) - \frac{\cos(1/y)}{y} + x} \end{aligned}

$y^2\cos(1/y) = 2x + 2y$

\begin{aligned} 2y\frac{dy}{dx}\cos(1/y) - y^2\sin(1/y)\frac{-\frac{dy}{dx}}{y^2} &= 2 + 2\frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{2}{2y\cos(1/y) + \sin(1/y) - 2} \end{aligned}

$\phi^{1/2} + r^{1/2} = 1$

\begin{aligned} \frac{1}{2}\phi^{-1/2} + \frac{1}{2}r^{-1/2}\frac{dr}{d\phi} &= 0 \\ \frac{dr}{d\phi} &= -\frac{\phi^{-1/2}}{r^{-1/2}} \end{aligned}

$r - 2\sqrt{\phi} = \frac{3}{2}\phi^{2/3} + \frac{4}{3}\phi^{3/4}$

\begin{aligned} \frac{dr}{d\phi} - \phi^{-1/2} &= \phi^{-1/3} + \phi^{-1/4} \\ \frac{dr}{d\phi} &= \phi^{-1/2} + \phi^{-1/3} + \phi^{-1/4} \end{aligned}

$\sin(r\phi) = \frac{1}{2}$

\begin{aligned} \cos(r\phi)\cdot (\frac{dr}{d\phi}\cdot \phi + r) &= 0 \\ \frac{dr}{d\phi} &= -\frac{r}{\phi} \end{aligned}

$\cos r + \cot \phi = r\phi$

\begin{aligned} -\sin r \frac{dr}{d\phi} - \csc^2 \phi &= \phi \frac{dr}{d\phi} + r \\ \frac{dr}{d\phi} &= \frac{-\csc^2\phi - r}{\phi + \sin r} \end{aligned}

$x^2 + y^2 = 1$

2x + 2y\frac{dy}{dx} = 0 \\ \frac{dy}{dx} = -\frac{x}{y}

\frac{d}{dx}(-\frac{x}{y}) = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{1}{y} - \frac{x^2}{y^3}

$x^{2/3} + y^{2/3} = 1$

\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \\ \frac{dy}{dx} = -(\frac{x}{y})^{-1/3}

\frac{d}{dx}(-(\frac{x}{y})^{-1/3}) = \frac{1}{3} \cdot \frac{y - x\frac{dy}{dx}}{y^2} \\ \frac{d^2y}{dx^2} = \frac{y + x(\frac{x}{y})^{-1/3}}{3y^2}

$y^2 = x^2 + 2x$

2y\frac{dy}{dx} = 2x + 2 \\ \frac{dy}{dx} = \frac{x + 1}{y}

\frac{d^2y}{dx^2} = \frac{y - (x + 1)\frac{dy}{dx}}{y^2} = \frac{y^2 - (x + 1)^2}{y^3}

$y^2 - 2x = 1 - 2y$

2y\frac{dy}{dx} - 2 = -2\frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{y + 1}

\frac{d^2y}{dx^2} = \frac{- \frac{dy}{dx}}{(y + 1)^2} = -\frac{1}{(y + 1)^3}