H1复合函数，链式法则 自变量\(x,y\)的一个函数\(u\)常常由形式 $$ u=f(\xi, \eta, \cdots) $$ 表出，其中\(f\)的自变量\(\xi, \eta, \cdots\)本身又是\(x\)和\(y\)的函数 $$ \xi=\varphi(x,y), \qquad \eta = \psi(x,y), \cdots $$ 于是，我们称 $$ u=f(\xi, \eta, \cdots)=f(\varphi(x,y), \psi(x,y), \cdots) = F(x,y) $$ 是\(x\)和\(y\)的一个复合函数。它也是\((x,y)\)的可微函数，它的偏导数由公式 $$ F_x = f_{\xi}\varphi_x + f_{\eta}\psi_x + \cdots, $$ $$ F_y = f_{\xi}\varphi_y + f_{\eta}\psi_y \cdots, $$ 可简写成 $$ u_x=u_{\xi}\xi_x + u_{\eta} \eta_x + \cdots, $$ $$ u_y=u_{\xi}\xi_y + u_{\eta} \eta_y + \cdots, $$ 这样，为了求出关于\(x\)的偏导数，我们必须首先求出复合函数关于每一个变量\(\xi, \eta,\cdots\)的导数，再把这个导数中的每一个乘以对应变量关于\(x\)的导数，并将所有的乘积相加。 H1例一 $$ u=F(x,y) = \log(x^4+y^4) \cdot arc \sin\sqrt{1-x^2-y^2} $$ 令\(\xi = \log(x^4+y^4), \eta = \sqrt{1-x^2-y^2}\)，那么可表示成复合函数 $$ u=f(\xi, \eta) = \xi \cdot arc \sin \eta $$ 求偏导数\(u_x, u_y\)，如下： $$ u_x = u_{\xi}\xi_x + u_{\eta} \eta_x=\frac{\partial u}{\partial \xi} \cdot \frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta} \cdot \frac{\partial \eta}{\partial x} $$ $$ =arc \sin \eta \cdot \frac{\partial \xi}{\partial x} + \frac{\xi}{\sqrt{1-\eta^2}} \cdot \frac{\partial \eta}{\partial x} $$ $$ u_y = u_{\xi}\xi_y + u_{\eta} \eta_y=\frac{\partial u}{\partial \xi} \cdot \frac{\partial \xi}{\partial y} + \frac{\partial u}{\partial \eta} \cdot \frac{\partial \eta}{\partial y} $$ $$ =arc \sin \eta \cdot \frac{\partial \xi}{\partial y} + \frac{\xi}{\sqrt{1-\eta^2}} \cdot \frac{\partial \eta}{\partial y} $$ 由于函数\(\xi, \eta\)可用复合函数来表示： $$ \xi = \log(x^4 + y^4) \implies \xi = f(\gamma) = \log \gamma, \gamma = x^4 + y^4 $$ $$ \eta = \sqrt{1-x^2-y^2} \implies \eta = f(\delta) = \sqrt{\delta}, \delta = 1-x^2-y^2 $$ 所以： $$ \frac{\partial \xi}{\partial x} = f_{\gamma}\gamma_x=\frac{\partial \xi}{\partial \gamma} \cdot \frac{\partial \gamma}{\partial x} = \frac{1}{\gamma} \cdot 4x^3 = \frac{4x^3}{x^4+y^4} $$ $$ \frac{\partial \xi}{\partial y} = f_{\gamma}\gamma_y=\frac{\partial \xi}{\partial \gamma} \cdot \frac{\partial \gamma}{\partial y} = \frac{1}{\gamma} \cdot 4y^3 = \frac{4y^3}{x^4+y^4} $$ $$ \frac{\partial \eta}{\partial x} = f_{\delta} \delta_x = \frac{\partial \eta}{\partial \delta} \cdot \frac{\partial \delta}{\partial x} = \frac{1}{2} \frac{1}{\sqrt{\delta}}(-2x)=-\frac{x}{\sqrt{1-x^2-y^2}} $$ $$ \frac{\partial \eta}{\partial y} = f_{\delta} \delta_y = \frac{\partial \eta}{\partial \delta} \cdot \frac{\partial \delta}{\partial y} = \frac{1}{2} \frac{1}{\sqrt{\delta}}(-2y)=-\frac{y}{\sqrt{1-x^2-y^2}} $$ 所以： $$ u_x=arc \sin \eta \cdot \frac{\partial \xi}{\partial x} + \frac{\xi}{\sqrt{1-\eta^2}} \cdot \frac{\partial \eta}{\partial x} $$ $$ =\frac{4x^3}{x^4+y^4} arc\sin \sqrt{1-x^2-y^2} + \frac{\log(x^4+y^4)}{\sqrt{1-(1-x^2-y^2)}} (-\frac{x}{\sqrt{1-x^2-y^2}}) $$ $$ =\frac{4x^3 arc\sin \sqrt{1-x^2-y^2} }{x^4+y^4} - \frac{x\log(x^4+y^4)}{\sqrt{(x^2+y^2)(1-x^2-y^2)}} $$ $$ u_y=arc \sin \eta \cdot \frac{\partial \xi}{\partial y} + \frac{\xi}{\sqrt{1-\eta^2}} \cdot \frac{\partial \eta}{\partial y} $$ $$ =\frac{4y^3arc\sin \sqrt{1-x^2-y^2}}{x^4+y^4}-\frac{y\log(x^4+y^4)}{\sqrt{(x^2+y^2)(1-x^2-y^2)}} $$ H1例二 $$ \omega=F(x,y)=\frac{1}{\sqrt{(x^2+y^2+2xy\cos z)}} $$ 令\(\xi=x^2+y^2+2xy\cos z\) 能表示成复合函数的形式： $$ \omega = f(\xi) = \frac{1}{\sqrt{\xi}} $$ $$ \omega_x=\omega_{\xi}\xi_x=\frac{\partial \omega}{\partial \xi}\cdot \frac{\partial \xi}{\partial x}= -\frac{1}{2}\frac{1}{\xi^{3/2}} (2x + 2y\cos z) = -\frac{x+y\cos z}{(x^2+y^2+2xy\cos z)^{3/2}} $$ $$ \omega_y=\omega_{\xi}\xi_y=\frac{\partial \omega}{\partial \xi}\cdot \frac{\partial \xi}{\partial y}= -\frac{1}{2}\frac{1}{\xi^{3/2}} (2y + 2x\cos z) = -\frac{y+x\cos z}{(x^2+y^2+2xy\cos z)^{3/2}} $$ $$ \omega_z=\omega_{\xi}\xi_y=\frac{\partial \omega}{\partial \xi}\cdot \frac{\partial \xi}{\partial z}= -\frac{1}{2}\frac{1}{\xi^{3/2}} (2xy\sin z) = \frac{xy\sin z}{(x^2+y^2+2xy\cos z)^{3/2}} $$
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