17.设 =varphi (x+y,(x)^2), 且φ具有二阶连续偏导数,求 dfrac (partial x)(partial x) ,dfrac ({a)^2z}(partial {x)^2} ,dfrac ({a)^2z}(partial xpartial y) →(a)

步骤 1：求 $\dfrac {\partial z}{\partial x}$
根据链式法则，我们有：
$$\dfrac {\partial z}{\partial x} = \dfrac {\partial \varphi}{\partial (x+y)} \cdot \dfrac {\partial (x+y)}{\partial x} + \dfrac {\partial \varphi}{\partial (x^2)} \cdot \dfrac {\partial (x^2)}{\partial x}$$
$$= \varphi_1' \cdot 1 + \varphi_2' \cdot 2x$$
$$= \varphi_1' + 2x\varphi_2'$$
步骤 2：求 $\dfrac {{\partial }^{2}z}{\partial {x}^{2}}$
对 $\dfrac {\partial z}{\partial x}$ 再次求偏导，我们有：
$$\dfrac {{\partial }^{2}z}{\partial {x}^{2}} = \dfrac {\partial}{\partial x}(\varphi_1' + 2x\varphi_2')$$
$$= \dfrac {\partial \varphi_1'}{\partial x} + 2\varphi_2' + 2x\dfrac {\partial \varphi_2'}{\partial x}$$
$$= \varphi_{11}'' \cdot 1 + \varphi_{12}'' \cdot 2x + 2\varphi_2' + 2x(\varphi_{21}'' \cdot 1 + \varphi_{22}'' \cdot 2x)$$
$$= \varphi_{11}'' + 2x\varphi_{12}'' + 2\varphi_2' + 2x\varphi_{21}'' + 4x^2\varphi_{22}''$$
步骤 3：求 $\dfrac {{\partial }^{2}z}{\partial x\partial y}$
对 $\dfrac {\partial z}{\partial x}$ 求关于 $y$ 的偏导，我们有：
$$\dfrac {{\partial }^{2}z}{\partial x\partial y} = \dfrac {\partial}{\partial y}(\varphi_1' + 2x\varphi_2')$$
$$= \dfrac {\partial \varphi_1'}{\partial y} + 2x\dfrac {\partial \varphi_2'}{\partial y}$$
$$= \varphi_{11}'' \cdot 1 + \varphi_{12}'' \cdot 2x$$
$$= \varphi_{11}'' + 2x\varphi_{12}''$$