Question

12设+z=f(x+y^2)+,x2-y),且f具有二阶连续偏导数,求02z(^3z)/(xy)

Answer 1

问：12设+z=f(x+y^2)+,x2-y),且f具有二阶连续偏导数,求02z(^3z)/(xy)

答：您好亲亲，首先，我们需要求出 $z$ 对 $x$ 和 $y$ 的一阶偏导数：$$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} [f(x+y^2) + x^2 - y] = f_x(x+y^2) + 2x$$$$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} [f(x+y^2) + x^2 - y] = 2y - f_{xy}(x+y^2)$$然后，我们对 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 再次求偏导数，得到二阶偏导数：$$\frac{\partial^2 z}{\partial x^2} = f_{xx}(x+y^2) + 2$$$$\frac{\partial^2 z}{\partial y^2} = 2 - f_{xxy}(x+y^2)$$$$\frac{\partial^2 z}{\partial x\partial

答：$$\frac{\partial^2 z}{\partial x\partial y} = -f_{xy}(x+y^2)$$

问：能不能用数学语言表示一下 这个frac啥的我实在看不懂

答：现在，我们可以开始求 $0.2z\left(\frac{\partial^3 z}{\partial x\partial y\partial z} \right)_{xy}$。按照题目给出的公式，有：$$0.2z\left(\frac{\partial^3 z}{\partial x\partial y\partial z} \right)_{xy} = 0.2z\left(\frac{\partial^3 z}{\partial x\partial z\partial y} \right)_{xy} = 0.2\frac{\partial z}{\partial y}\left(\frac{\partial^2 z}{\partial x\partial z} \right)_y$$

答：亲亲，用不了数学语言，转换不了

答：题目中给出的方程可以写成 $z=f(x+y^2,x^2-y)$，则有：$$\frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}=f_{1}+f_{2}(2x)\\\frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}=2f_{2}y\\\frac{\partial^2 z}{\partial x^2}=\frac{\partial^2 f}{\partial x^2}+2\frac{\partial^2 f}{\partial u^2}\left(\frac{\partial u}{\partial x}\right)^2+2\frac{\partial^2 f}{\partial u\partial x}\frac{\partial u}{\partial x＝

答：f_{11}+4xf_{12}+4f_{22}x^2\\\frac{\partial^2 z}{\partial y^2}=2f_{22}\\\frac{\partial^2 z}{\partial x\partial y}=\frac{\partial^2 f}{\partial u^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+\frac{\partial^2 f}{\partial u\partial y}\frac{\partial u}{\partial x}=2yf_{12}+2f_{22}x$$

答：代入题目中的式子：$$\begin{aligned}&\frac{\partial z^3}{\partial xy}=\frac{\partial}{\partial x}\left(z^2\frac{\partial z}{\partial y}\right)-\frac{\partial}{\partial y}\left(z^2\frac{\partial z}{\partial x}\right)\\

问：行吧

答：=&\frac{\partial}{\partial x}\left((f_{1}+f_{2}(2x))^2\cdot2f_{2}y\right)-\frac{\partial}{\partial y}\left((f_{1}+f_{2}(2x))^2\cdot(f_{11}+4xf_{12}+4f_{22}x^2)\right)\\=&2(f_{11}+4xf_{12}+6f_{22}x^2)f_{2}^2-2(f_{1}+2f_{2}x)^2f_{22}\end{aligned}$$

答：所以最终的答案是$$\frac{\partial^3 z}{\partial xy^2}=\frac{\partial^3 z}{\partial y\partial x^2}=0\\\frac{\partial^3 z}{\partial x^3}=6f_{12}f_{2}^2+24f_{22}xf_{2}^2+4f_{11}f_{2}f_{22}+12xf_{1}f_{2}f_{22}-4(f_{1}+2f_{2}x)^2f_{222}\\

答：\frac{\partial^3 z}{\partial y^3}=8f_{22}^3\\\frac{\partial^3 z}{\partial x^2\partial y}=2f_{12}(6f_{2}^2+2f_{1}f_{22})+8xf_{2}f_{22}(3f_{2}+f_{1})-4(f_{1}+2f_{2}x)f_{222}f_{22}\\\frac{\partial^3 z}{\partial xy^2}=\frac{\partial^3 z}{\partial y\partial x^2}=0\\$$

答：亲亲，信息过程如上所述哈