我们需要分别对$x$和$y$求偏导数,然后代入$(2,2)$计算。
$$
\frac{\partial z}{\partial x} = \frac{1}{x^2+y^3} \cdot 2x = \frac{2x}{x^2+y^3}
$$
$$
\frac{\partial z}{\partial y} = \frac{1}{x^2+y^3} \cdot 3y^2 = \frac{3y^2}{x^2+y^3}
$$
将$(x,y)=(2,2)$代入上式,得到:
$$
\frac{\partial z}{\partial x}\bigg|_{(2,2)} = \frac{2\cdot 2}{2^2+2^3} = \frac{4}{12} = \frac{1}{3}
$$
$$
\frac{\partial z}{\partial y}\bigg|_{(2,2)} = \frac{3\cdot 2^2}{2^2+2^3} = \frac{12}{12} = 1
$$
因此,$z=ln(x^2+y^3)$在$(2,2)$处的一阶偏导数分别为$\frac{\partial z}{\partial x}\bigg|_{(2,2)}=\frac{1}{3}$和$\frac{\partial z}{\partial y}\bigg|_{(2,2)}=1$。