题目
设=f((x)^3+(y)^2), 其中f具有二阶连续偏导数,则 dfrac ({partial )^2z}(partial {y)^2}=
题目解答
答案
方程两边同时对
求偏导得：
由此可得本题答案
解析
步骤 1：对$z=f({x}^{3}+{y}^{2})$关于$y$求偏导
根据链式法则，我们有
$$\dfrac {\partial z}{\partial y} = f'({x}^{3}+{y}^{2}) \cdot \dfrac {\partial }{\partial y}({x}^{3}+{y}^{2}) = f'({x}^{3}+{y}^{2}) \cdot 2y$$
步骤 2：对$\dfrac {\partial z}{\partial y}$关于$y$求偏导
再次应用链式法则，我们有
$$\dfrac {{\partial }^{2}z}{\partial {y}^{2}} = \dfrac {\partial }{\partial y} \left( f'({x}^{3}+{y}^{2}) \cdot 2y \right)$$
$$= 2y \cdot \dfrac {\partial }{\partial y} f'({x}^{3}+{y}^{2}) + f'({x}^{3}+{y}^{2}) \cdot \dfrac {\partial }{\partial y} (2y)$$
$$= 2y \cdot f''({x}^{3}+{y}^{2}) \cdot 2y + 2f'({x}^{3}+{y}^{2})$$
$$= 4y^2 f''({x}^{3}+{y}^{2}) + 2f'({x}^{3}+{y}^{2})$$
根据链式法则，我们有
$$\dfrac {\partial z}{\partial y} = f'({x}^{3}+{y}^{2}) \cdot \dfrac {\partial }{\partial y}({x}^{3}+{y}^{2}) = f'({x}^{3}+{y}^{2}) \cdot 2y$$
步骤 2：对$\dfrac {\partial z}{\partial y}$关于$y$求偏导
再次应用链式法则，我们有
$$\dfrac {{\partial }^{2}z}{\partial {y}^{2}} = \dfrac {\partial }{\partial y} \left( f'({x}^{3}+{y}^{2}) \cdot 2y \right)$$
$$= 2y \cdot \dfrac {\partial }{\partial y} f'({x}^{3}+{y}^{2}) + f'({x}^{3}+{y}^{2}) \cdot \dfrac {\partial }{\partial y} (2y)$$
$$= 2y \cdot f''({x}^{3}+{y}^{2}) \cdot 2y + 2f'({x}^{3}+{y}^{2})$$
$$= 4y^2 f''({x}^{3}+{y}^{2}) + 2f'({x}^{3}+{y}^{2})$$