类似地，

$$\begin{aligned}\frac{\partial f}{\partial y} &= \frac{\partial}{\partial y}(3x^2) + \frac{\partial}{\partial y}(2xy) + \frac{\partial}{\partial y}(y^3) \ &= 0 + 2x \frac{d}{dy}(y) + \frac{d}{dy}(y^3) = 2x + 3y^2\end{aligned}$$

(b) $$f(x, y) = (y^2 + x) e^{-xy}$$

使用乘积法则，对 $x$ 求导，将 $y$ 视为常数，得到

$$\begin{aligned}\frac{\partial f}{\partial x} &= (e^{-xy}) \frac{\partial}{\partial x} (y^2 + x) + (y^2 + x) \frac{\partial}{\partial x} (e^{-xy}) \ &= (e^{-xy})(1) + (y^2 + x)(-ye^{-xy}) \ &= (1 - y^3 - xy)e^{-xy}\end{aligned}$$

类似地，

$$\begin{aligned} \frac{\partial f}{\partial y} &= (e^{-xy}) \frac{\partial}{\partial y} (y^2 + x) + (y^2 + x) \frac{\partial}{\partial y} (e^{-xy}) \ &= (e^{-xy})(2y) + (y^2 + x)(-xe^{-xy}) \ &= (2y - xy^2 - x^2)e^{-xy} \end{aligned}$$

例 9.25 求 $\partial f/\partial x$ 和 $\partial f/\partial y$ 当 $f(x, y)$ 为

(a) $$xy^2 + 3xy - x + 2$$      (b) $\sin(x^2 - 3y)$

解

解 (a) 取 $f(x, y) = xy^2 + 3xy - x + 2$ 并对 $x$ 求偏导，保持 $y$ 不变，得到

$$\frac{\partial f}{\partial x} = f_x = y^2 + 3y - 1$$

对 $y$ 求偏导，保持 $x$ 不变，得到

$$\frac{\partial f}{\partial y} = f_y = 2xy + 3x$$

(b) 取 $f(x, y) = \sin(x^2 - 3y)$ 并应用复合函数求导法则，我们得到

$$\begin{aligned}\frac{\partial f}{\partial x} &= \cos(x^2 - 3y) \frac{\partial}{\partial x}(x^2 - 3y) = \cos(x^2 - 3y) 2x \ &= 2x \cos(x^2 - 3y)\end{aligned}$$

和

$$\frac{\partial f}{\partial y} = \cos(x^2 - 3y) \frac{\partial}{\partial y}(x^2 - 3y) = -3 \cos(x^2 - 3y)$$