设fxy＝x³+3y²+x²+y²xy求f₁₂xy

首先求$f(x,y)$的一阶偏导数： $$ \begin{aligned} \frac{\partial f}{\partial x} &= 3x^2 + \frac{2x}{y} + \frac{2y}{x^2+y^2}\ \frac{\partial f}{\partial y} &= 6y + \frac{2y}{x^2+y^2} + \frac{2x}{y^2} \end{aligned} $$ 然后对$\frac{\partial f}{\partial x}$求一次偏导数，得到二阶混合偏导数： $$ \frac{\partial^2f}{\partial x^2} = 6x - \frac{2x^3+6xy^2}{(x^2+y^2)^2} + \frac{2}{y^2} \qquad \frac{\partial^2f}{\partial y\partial x} = \frac{2}{x} - \frac{4xy}{(x^2+y^2)^2} + \frac{2}{y^2} $$ 最后对$\frac{\partial f}{\partial y}$求一次偏导数，得到二阶混合偏导数： $$ \frac{\partial^2f}{\partial y\partial x} = \frac{2}{x} - \frac{4xy}{(x^2+y^2)^2} + \frac{2}{y^2} \qquad \frac{\partial^2f}{\partial y^2} = 6 + \frac{2x^2-2y^2}{(x^2+y^2)^2} $$ 所以，$f_{12}''(x,y) = \frac{\partial^2f}{\partial y\partial x} = \frac{\partial^2f}{\partial x\partial y} = \frac{2}{x} - \frac{4xy}{(x^2+y^2)^2} + \frac{2}{y^2}$