Differentiating $f_x$ 关于 $x$ gives

$$\begin{aligned} f_{xx} &= \frac{\partial}{\partial x} \left( 2x \frac{\partial f}{\partial s} + y \frac{\partial f}{\partial t} \right) \ &= 2 \frac{\partial f}{\partial s} + 2x \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial s} \right) + y \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial t} \right) \quad (\text{using the product rule}) \end{aligned}$$

如上所述，反复使用链式法则可得

$$\begin{aligned} f_{xx} &= 2 \frac{\partial f}{\partial s} + 2x \left[ \frac{\partial}{\partial s} \left( \frac{\partial f}{\partial s} \right) \frac{\partial s}{\partial x} + \frac{\partial}{\partial t} \left( \frac{\partial f}{\partial s} \right) \frac{\partial t}{\partial x} \right] + y \left[ \frac{\partial}{\partial s} \left( \frac{\partial f}{\partial t} \right) \frac{\partial s}{\partial x} + \frac{\partial}{\partial t} \left( \frac{\partial f}{\partial t} \right) \frac{\partial t}{\partial x} \right] \ &= 2f_s + 2x(2xf_{ss} + yf_{st}) + y(2xf_{ts} + yf_{tt}) \end{aligned}$$

which, 在假设……的基础上 $f_{st} = f_{ts}$ , gives

$$f_{xx} = 2f_s + 4x^2f_{ss} + y^2f_{tt} + 4xyf_{st} \quad (9.26)$$

按照类似的步骤, 我们可以确定 $f_{yy}$ . Differentiating $f_y$ 关于；相对于 $y$ gives

$$\begin{aligned} f_{yy} &= \frac{\partial}{\partial y} (-2yf_s + xf_t) \ &= -2f_s - 2y \frac{\partial}{\partial y} (f_s) + x \frac{\partial}{\partial y} (f_t) \ &= -2f_s - 2y \left[ \frac{\partial}{\partial s} (f_s) \frac{\partial s}{\partial y} + \frac{\partial}{\partial t} (f_s) \frac{\partial t}{\partial y} \right] + x \left[ \frac{\partial}{\partial s} (f_t) \frac{\partial s}{\partial y} + \frac{\partial}{\partial t} (f_t) \frac{\partial t}{\partial y} \right] \ &= -2f_s - 2y(-2yf_{ss} + xf_{st}) + x(-2yf_{ts} + xf_{tt}) \end{aligned}$$

giving

$$f_{yy} = -2f_s + 4y^2f_{ss} + x^2f_{tt} - 4xyf_{st} \quad (9.27)$$

Adding (9.26) and (9.27), 我们得到

$$\begin{aligned} f_{xx} + f_{yy} &= 4(x^2 + y^2)f_{ss} + (x^2 + y^2)f_{tt} \ &= (x^2 + y^2)(4f_{ss} + f_{tt}) \ &= \sqrt{(x^2 - y^2)^2 + 4x^2y^2} [4f_{ss} + f_{tt}] \end{aligned}$$

从而得到所需的结果。

$$f_{xx} + f_{yy} = \sqrt{(s^2 + 4t^2)(4f_{ss} + f_{tt})}$$

9.6.8 Exercises

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