Similarly $\frac{\partial f}{\partial y}$ at $(1, 2)$ 由下式给出

$$\begin{aligned}\frac{\partial f}{\partial y} &= \lim_{\Delta y \rightarrow 0} \frac{[1 + 3(2 + \Delta y) + (2 + \Delta y)^2] - 11}{\Delta y} \ &= \lim_{\Delta y \rightarrow 0} \frac{3\Delta y + 4\Delta y + \Delta y^2}{\Delta y} = 7\end{aligned}$$

Example 9.23用定义（第一性原理）求下列函数的一阶偏导数： $f(x, y) = y \sin x$ 在一般点处 $(x, y)$ .

Solution Since $y$ 与……无关 $x$

$$\begin{aligned}\frac{\partial f}{\partial x} &= \lim_{\Delta x \rightarrow 0} \frac{y \sin(x + \Delta x) - y \sin x}{\Delta x} = y \lim_{\Delta x \rightarrow 0} \frac{\sin(x + \Delta x) - \sin x}{\Delta x} \ \frac{\sin(x + \Delta x) - \sin x}{\Delta x} &= \frac{2 \cos \frac{1}{2}(2x + \Delta x) \sin \frac{1}{2}\Delta x}{\Delta x} = \cos(x + \frac{1}{2}\Delta x) \frac{\sin \frac{1}{2}\Delta x}{\frac{1}{2}\Delta x} \ \text{As } \Delta x \rightarrow 0, \cos(x + \frac{1}{2}\Delta x) &\rightarrow \cos x \text{ and } \frac{\sin \frac{1}{2}\Delta x}{\frac{1}{2}\Delta x} \rightarrow 1\end{aligned}$$

(参见第 7.8.1). Thus

$$\frac{\partial f}{\partial x} = y \cos x$$

$$\begin{aligned}\text{Similarly } \frac{\partial f}{\partial y} &= \lim_{\Delta y \rightarrow 0} \frac{(y + \Delta y) \sin x - y \sin x}{\Delta y} \ &= \lim_{\Delta y \rightarrow 0} \frac{\Delta y \sin x}{\Delta y} = \sin x\end{aligned}$$

Example 9.24Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ where $f(x, y)$ 由下式给出

(a) $3x^2 + 2xy + y^3$ (b) $(y^2 + x)e^{-xy}$

Solution (a) $f(x, y) = 3x^2 + 2xy + y^3$

求 $\frac{\partial f}{\partial x}$ , 我们求导 $f(x, y)$ 关于 $x$ regarding $y$ 作为一个常数. 于是我们得到

$$\begin{aligned}\frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}(3x^2) + \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial x}(y^3) \ &= 3 \frac{d}{dx}(x^2) + 2y \frac{d}{dx}(x) + 0 \quad (\text{Note: term in brackets involves } x \text{ only}) \ &= 6x + 2y\end{aligned}$$