多元复合函数的求导法则

这里我们推导多元复合函数的求导法则。需要注意的是，多元复合函数，有多少个中间变量，就应该有多少项。推导的方法与一元复合函数差不多。

1，一般情形，两个中间变量，两个自变量：设 $z = f (u, v), u = g (x, y), v = h (x, y)$ ， $f, g, h$ 可微，则

$\begin{aligned} \frac{\partial f}{\partial x} & = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} = f_{u} \cdot g_{x} + f_{v} \cdot h_{x} \\ \frac{\partial f}{\partial y} & = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} = f_{u} \cdot g_{y} + f_{v} \cdot h_{y} \end{aligned}$

2，两个中间变量，一个自变量：设 $z = f (u, v), u = g (t), v = h (t)$ ，则

$\frac{d z}{d t} = \frac{\partial f}{\partial u} \cdot \frac{d u}{d t} + \frac{\partial f}{\partial v} \cdot \frac{d v}{d t} = f_{u} \cdot g^{'} (t) + f_{v} \cdot h^{'} (t)$

3，函数中既有自变量，又有中间变量，自变量只有一个： $z = f (t, u, v), u = g (t), v = h (t)$ ，那么

$\frac{d z}{d t} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} \cdot \frac{d u}{d t} + \frac{\partial f}{\partial v} \cdot \frac{d v}{d t} = f_{t} + f_{x} \cdot g^{'} (t) + f_{y} \cdot h^{'} (t)$

4，函数中既有自变量，又有中间变量，自变量有两个： $w = f (x, y, z), z = g (x, y)$ ，那么

$\begin{aligned} \frac{\partial w}{\partial x} & = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z} \cdot \frac{\partial g}{\partial x} = f_{x} + f_{z} \cdot g_{x} \\ \frac{\partial w}{\partial y} & = \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} \cdot \frac{\partial g}{\partial y} = f_{y} + f_{z} \cdot g_{y} \end{aligned}$

我们仅仅证明第二种情况，其它情形类似，只是稍微烦琐些。

2 的证明：因为

$\frac{d z}{d t} = lim_{Δ t \to 0} \frac{Δ z}{Δ t}$

因为 $f$ 可微，所以 $Δ z = f_{u} \cdot Δ u + f_{v} \cdot Δ v + o (ρ)$ 这里 $ρ = \sqrt{Δ^{2} u + Δ^{2} v}$ 。所以可以得到 $\frac{Δ z}{Δ t} = f_{u} \cdot \frac{Δ u}{Δ t} + f_{v} \cdot \frac{Δ v}{Δ t} + \frac{o (ρ)}{Δ t}$

又因为 $g, h$ 可微，并且 $lim_{Δ u \to 0} \frac{o (ρ)}{Δ u} = 0, lim_{Δ v \to 0} \frac{o (ρ)}{Δ u} = 0$ 所以 $lim_{Δ t \to 0} \frac{o (ρ)}{Δ t} = lim_{Δ t \to 0} \frac{o (ρ) / Δ u}{Δ t / Δ u} = 0$ 这是因为分母的极限为 $1 / g^{'} (t)$ 。

所以当 $Δ t \to 0$ 时， $lim_{Δ t \to 0} \frac{Δ z}{Δ t} = lim_{Δ \to 0} (f_{u} \cdot \frac{Δ u}{Δ t} + f_{v} \cdot \frac{Δ v}{Δ t} + \frac{o (ρ)}{Δ t}) = f_{u} \cdot \frac{d u}{d t} + f_{v} \cdot \frac{d v}{d t}$

我们来看几个例题。

例1，设 $z = e^{u^{2} + 2 u v - v^{2}}, u = 2 x - y, v = x + 2 y$ ，求 $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$ 。

解：这是复合函数的第一种情况，所以 $\begin{aligned} \frac{\partial z}{\partial x} & = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ = (2 u + 2 v) e^{u^{2} + 2 u v - v^{2}} \cdot 2 + (2 u - 2 v) e^{u^{2} + 2 u v - v^{2}} \cdot 1 \\ = e^{u^{2} + 2 u v - v^{2}} (4 u + 4 v + 2 u - 2 v) \\ = (6 u + 2 v) e^{u^{2} + 2 u v - v^{2}} \end{aligned}$

$\begin{aligned} \frac{\partial z}{\partial y} & = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \\ = (2 u + 2 v) e^{u^{2} + 2 u v - v^{2}} \cdot (- 1) + (2 u - 2 v) e^{u^{2} + 2 u v - v^{2}} \cdot 2 \\ = e^{u^{2} + 2 u v - v^{2}} (- 2 u - 2 v + 4 u - 4 v) \\ = (2 u - 6 v) e^{u^{2} + 2 u v - v^{2}} \end{aligned}$

例2，设 $z = \sin (x y) + e^{x + 2 y}, x = \sin t, y = \cos t$ ，求 $\frac{d z}{d t}$ 。

解：这是第二种情况，所以

$\begin{aligned} \frac{d z}{d t} & = \frac{\partial f}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial f}{\partial y} \cdot \frac{d y}{d t} \\ = y \cos (x y) \cdot \cos t + x \cos (x y) \cdot (- \sin t) \\ = \cos (x y) (\cos^{2} t - \sin^{2} t) = \cos 2 t \cdot \cos (x y) \end{aligned}$

例3，设 $z = \ln (t + x^{2} + y^{2}), x = \sin t, y = \cos t$ ，求 $\frac{d z}{d t}$ 。

解：这是第三种情形。所以

$\begin{aligned} \frac{d z}{d t} & = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial f}{\partial y} \cdot \frac{d y}{d t} \\ = \frac{1}{t + x^{2} + y^{2}} + \frac{2 x}{t + x^{2} + y^{2}} \cdot \cos t + \frac{2 y}{t + x^{2} + y^{2}} \cdot (- \sin t) \\ = \frac{1}{t + x^{2} + y^{2}} (1 + 2 \sin t \cos t - 2 \cos t \sin t) \\ = \frac{1}{t + x^{2} + y^{2}} \end{aligned}$

例4，设 $w = e^{x^{2} + y^{2} + z^{2}}, z = \ln (x^{2} + 2 x y - y^{2})$ ，求 $\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}$ 。

解：这是第四种情形的复合函数。我们有

$\begin{aligned} \frac{\partial w}{\partial x} & = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial x} \\ = 2 x e^{x^{2} + y^{2} + z^{2}} + 2 z e^{x^{2} + y^{2} + z^{2}} \cdot \frac{2 x + 2 y}{x^{2} + 2 x y - y^{2}} \end{aligned}$

$\begin{aligned} \frac{\partial w}{\partial y} & = \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} \cdot \frac{\partial z}{\partial y} \\ = 2 y e^{x^{2} + y^{2} + z^{2}} + 2 z e^{x^{2} + y^{2} + z^{2}} \cdot \frac{2 x - 2 y}{x^{2} + 2 x y - y^{2}} \end{aligned}$