五、复合函数求导法则的其他应用
（一）变量替换下方程的变形
多元函数的复合函数求导法则的重要应用在于：作变量替换时，求函数在新变量下的偏导数，有时通过变量替换可将某些偏微分方程（未知函数及其偏导数所满足的方程）化简。
【例8.9】 设 \(u = u(x,y)\) 有二阶连续偏导数，证明：在极坐标变换 \(x = r\cos \theta\)，\(y = r\sin \theta\) 下有
\[\frac {\partial^ {2} u}{\partial x ^ {2}} + \frac {\partial^ {2} u}{\partial y ^ {2}} = \frac {\partial^ {2} u}{\partial r ^ {2}} + \frac {1}{r} \frac {\partial u}{\partial r} + \frac {1}{r ^ {2}} \frac {\partial^ {2} u}{\partial \theta^ {2}}.\]【证明】 利用复合函数求导公式，有
\[\frac {\partial u}{\partial r} = \frac {\partial u}{\partial x} \frac {\partial x}{\partial r} + \frac {\partial u}{\partial y} \frac {\partial y}{\partial r} = \cos \theta \frac {\partial u}{\partial x} + \sin \theta \frac {\partial u}{\partial y}, \tag {*}\] \[\frac {\partial u}{\partial \theta} = \frac {\partial u}{\partial x} \frac {\partial x}{\partial \theta} + \frac {\partial u}{\partial y} \frac {\partial y}{\partial \theta} = - r \sin \theta \frac {\partial u}{\partial x} + r \cos \theta \frac {\partial u}{\partial y},\] \[\frac {\partial^ {2} u}{\partial r ^ {2}} = \cos \theta \frac {\partial}{\partial r} \left(\frac {\partial u}{\partial x}\right) + \sin \theta \frac {\partial}{\partial r} \left(\frac {\partial u}{\partial y}\right) = \cos^ {2} \theta \frac {\partial^ {2} u}{\partial x ^ {2}} + 2 \sin \theta \cos \theta \frac {\partial^ {2} u}{\partial x \partial y} + \sin^ {2} \theta \frac {\partial^ {2} u}{\partial y ^ {2}},\] \[\frac {\partial^ {2} u}{\partial \theta^ {2}} = - r \sin \theta \frac {\partial}{\partial \theta} \left(\frac {\partial u}{\partial x}\right) + r \cos \theta \frac {\partial}{\partial \theta} \left(\frac {\partial u}{\partial y}\right) - r \cos \theta \frac {\partial u}{\partial x} - r \sin \theta \frac {\partial u}{\partial y}.\]再对 \(\frac{\partial}{\partial\theta}\left(\frac{\partial u}{\partial x}\right)\) 及 \(\frac{\partial}{\partial\theta}\left(\frac{\partial u}{\partial y}\right)\) 用复合函数求导法及（*）式可得
\[\frac {\partial^ {2} u}{\partial \theta^ {2}} = (- r \sin \theta) ^ {2} \frac {\partial^ {2} u}{\partial x ^ {2}} - 2 r ^ {2} \sin \theta \cos \theta \frac {\partial^ {2} u}{\partial x \partial y} + (r \cos \theta) ^ {2} \frac {\partial^ {2} u}{\partial y ^ {2}} - r \frac {\partial u}{\partial r}.\]于是 \(\frac{\partial^2u}{\partial r^2} +\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2} = \frac{\partial^2u}{\partial x^2} +\frac{\partial^2u}{\partial y^2} -\frac{1}{r}\frac{\partial u}{\partial r},\) 即 \(\frac{\partial^2u}{\partial x^2} +\frac{\partial^2u}{\partial y^2} = \frac{\partial^2u}{\partial r^2} +\frac{1}{r}\frac{\partial u}{\partial r} +\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2}.\)
【注】 在极坐标变换 \(x = r\cos \theta ,y = r\sin \theta\) 下，拉普拉斯方程
\[\frac {\partial^ {2} u}{\partial x ^ {2}} + \frac {\partial^ {2} u}{\partial y ^ {2}} = 0 \quad \text {化 成} \quad \frac {\partial^ {2} u}{\partial r ^ {2}} + \frac {1}{r} \frac {\partial u}{\partial r} + \frac {1}{r ^ {2}} \frac {\partial^ {2} u}{\partial \theta^ {2}} = 0 .\]（二）多元函数问题转化为一元函数问题
多元函数问题常常转化为一元函数问题，通过复合函数求导法把一元函数的导数与多元函数的偏导数联系起来。如求二元函数 \(z = f(x, y)\) 在 \((x_0, y_0)\) 的二阶泰勒公式时，把二元函数的增量 \(f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0)\) 看成一元函数 \(\Phi(t) = f(x_0 + t\Delta x, y_0 + t\Delta y)\) 的增量 \(\Phi(1) - \Phi(0)\)。由一元函数的二阶泰勒公式：
\[\Phi (1) = \Phi (0) + \Phi^ {\prime} (0) + \frac {1}{2 !} \Phi^ {\prime \prime} (0) + \frac {1}{3 !} \Phi^ {(3)} (\theta) \quad (0 < \theta < 1),\]通过复合函数求导法求出 \(\varPhi^{\prime}(0),\varPhi^{\prime \prime}(0),\varPhi^{(3)}(\theta)\)，即得二元函数的二阶泰勒公式：
【定理8.11】 设 \(z = f(x,y)\) 在点 \(M_0(x_0,y_0)\) 的某邻域 \(U(M_0)\) 有直到3阶连续偏导数，\((x_0 + \Delta x, y_0 + \Delta y) \in U(M_0)\) 为任一点，则有二阶泰勒公式
\[\begin{aligned} f \left(x _ {0} + \Delta x, y _ {0} + \Delta y\right) &= f \left(x _ {0}, y _ {0}\right) + \left(\frac {\partial}{\partial x} \Delta x + \frac {\partial}{\partial y} \Delta y\right) f \left(x _ {0}, y _ {0}\right) \\ &\quad + \frac {1}{2 !} \left(\frac {\partial}{\partial x} \Delta x + \frac {\partial}{\partial y} \Delta y\right) ^ {2} f \left(x _ {0}, y _ {0}\right) + R _ {2} \\ &= f \left(x _ {0}, y _ {0}\right) + \frac {\partial f \left(x _ {0} , y _ {0}\right)}{\partial x} \Delta x + \frac {\partial f \left(x _ {0} , y _ {0}\right)}{\partial y} \Delta y \\ &\quad + \frac {1}{2} \left[ \frac {\partial^ {2} f \left(x _ {0} , y _ {0}\right)}{\partial x ^ {2}} \Delta x ^ {2} + 2 \frac {\partial^ {2} f (x _ {0} , y _ {0})}{\partial x \partial y} \Delta x \Delta y + \frac {\partial^ {2} f (x _ {0} , y _ {0})}{\partial y ^ {2}} \Delta y ^ {2} \right] + R _ {2}, \end{aligned}\]其中 \(R_{2} = \frac{1}{3!}\left(\frac{\partial}{\partial x}\Delta x + \frac{\partial}{\partial y}\Delta y\right)^{3}f(x_{0} + \theta \Delta x,y_{0} + \theta \Delta y)\quad (0 < \theta < 1)\) 称为拉格朗日余项，
\(R_{2} = o(\rho^{2})\) \((\rho = \sqrt{\Delta x^2 + \Delta y^2}\rightarrow 0)\) 称为皮亚诺余项。
这里 \(\left(\frac{\partial}{\partial x}\Delta x + \frac{\partial}{\partial y}\Delta y\right)^n f(x_0,y_0) = \sum_{k = 1}^n C_n^k\frac{\partial^n f(x_0,y_0)}{\partial x^k\partial y^{n - k}}\Delta x^k\Delta y^{n - k}\)，其中 \(C_n^k = \frac{n!}{k!(n - k)!}\)。
练习题
例题1
设 \(u = u(x, y)\) 有二阶连续偏导数，且满足拉普拉斯方程 \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\)。在极坐标变换 \(x = r\cos\theta, y = r\sin\theta\) 下，证明 \(u\) 满足的方程变为：
\[\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0.\]题目解答
由例8.9的结论，在极坐标变换下：
由于原方程 \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\)，代入即得：
\[\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0.\]例题2
设 \(z = f(x, y)\) 在点 \((1, 2)\) 的某邻域内有连续的一阶偏导数，且 \(f_x(1, 2) = 3\), \(f_y(1, 2) = -1\)。定义函数 \(\Phi(t) = f(1 + t\Delta x, 2 + t\Delta y)\)，其中 \(\Delta x = 0.1\), \(\Delta y = 0.2\)。求 \(\Phi'(0)\)。
题目解答
由复合函数求导法则：
代入 \(t = 0\)，得：
\[\Phi'(0) = f_x(1, 2) \cdot \Delta x + f_y(1, 2) \cdot \Delta y = 3 \times 0.1 + (-1) \times 0.2 = 0.3 - 0.2 = 0.1.\]例题3
设 \(u = u(x, y)\) 有二阶连续偏导数，且在极坐标变换 \(x = r\cos\theta, y = r\sin\theta\) 下，已知：
\[\frac{\partial u}{\partial r} = \cos\theta \frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}, \quad \frac{\partial u}{\partial \theta} = -r\sin\theta \frac{\partial u}{\partial x} + r\cos\theta \frac{\partial u}{\partial y}.\]求 \(\frac{\partial^2 u}{\partial r^2}\) 的表达式。
题目解答
对 \(\frac{\partial u}{\partial r}\) 再求偏导：
由于 \(\cos\theta\) 和 \(\sin\theta\) 与 \(r\) 无关，应用复合函数求导法则：
\[\frac{\partial^2 u}{\partial r^2} = \cos\theta \frac{\partial}{\partial r} \left( \frac{\partial u}{\partial x} \right) + \sin\theta \frac{\partial}{\partial r} \left( \frac{\partial u}{\partial y} \right).\]而
\[\frac{\partial}{\partial r} \left( \frac{\partial u}{\partial x} \right) = \frac{\partial^2 u}{\partial x^2} \frac{\partial x}{\partial r} + \frac{\partial^2 u}{\partial x \partial y} \frac{\partial y}{\partial r} = \cos\theta \frac{\partial^2 u}{\partial x^2} + \sin\theta \frac{\partial^2 u}{\partial x \partial y},\] \[\frac{\partial}{\partial r} \left( \frac{\partial u}{\partial y} \right) = \frac{\partial^2 u}{\partial y \partial x} \frac{\partial x}{\partial r} + \frac{\partial^2 u}{\partial y^2} \frac{\partial y}{\partial r} = \cos\theta \frac{\partial^2 u}{\partial x \partial y} + \sin\theta \frac{\partial^2 u}{\partial y^2}.\]代入得：
\[\frac{\partial^2 u}{\partial r^2} = \cos^2\theta \frac{\partial^2 u}{\partial x^2} + 2\sin\theta\cos\theta \frac{\partial^2 u}{\partial x \partial y} + \sin^2\theta \frac{\partial^2 u}{\partial y^2}.\]例题4
设 \(z = f(x, y)\) 在点 \((0, 0)\) 的某邻域内有直到三阶的连续偏导数，且 \(f(0, 0) = 1\), \(f_x(0, 0) = 2\), \(f_y(0, 0) = -1\), \(f_{xx}(0, 0) = 3\), \(f_{xy}(0, 0) = 0\), \(f_{yy}(0, 0) = 4\)。利用二元函数的二阶泰勒公式（带皮亚诺余项），求 \(f(0.1, -0.1)\) 的近似值。
题目解答
由定理8.11，二阶泰勒公式为：
其中 \(\rho = \sqrt{\Delta x^2 + \Delta y^2}\)。代入 \((x_0, y_0) = (0, 0)\), \(\Delta x = 0.1\), \(\Delta y = -0.1\)，及给定偏导数值：
\[f(0.1, -0.1) \approx 1 + 2 \times 0.1 + (-1) \times (-0.1) + \frac{1}{2} \left( 3 \times (0.1)^2 + 2 \times 0 \times 0.1 \times (-0.1) + 4 \times (-0.1)^2 \right).\]计算：
\[= 1 + 0.2 + 0.1 + \frac{1}{2} (0.03 + 0 + 0.04) = 1.3 + \frac{1}{2} \times 0.07 = 1.3 + 0.035 = 1.335.\]故 \(f(0.1, -0.1) \approx 1.335\)。