1.求下列函数的偏导数:(1)z=x^2y; (2)z=ycosx;(3)z=(1)/(sqrt(x^2)+y^(2)); (4)z=ln(x+y^2);(5)z=e^xy; (6)z=arctan(y)/(x);

以下是图片中各题的偏导数求解过程：

(1) $z = x^2 y$

• 对 $x$ 求偏导（将 $y$ 视为常数）：
  $z_x = \frac{\partial}{\partial x}(x^2 y) = y \cdot \frac{\partial}{\partial x}(x^2) = y \cdot 2x = 2xy$

• 对 $y$ 求偏导（将 $x$ 视为常数）：
  $z_y = \frac{\partial}{\partial y}(x^2 y) = x^2 \cdot \frac{\partial}{\partial y}(y) = x^2 \cdot 1 = x^2$

(2) $z = y \cos x$

• 对 $x$ 求偏导（将 $y$ 视为常数）：
  $z_x = \frac{\partial}{\partial x}(y \cos x) = y \cdot \frac{\partial}{\partial x}(\cos x) = y \cdot (-\sin x) = -y \sin x$

• 对 $y$ 求偏导（将 $x$ 视为常数）：
  $z_y = \frac{\partial}{\partial y}(y \cos x) = \cos x \cdot \frac{\partial}{\partial y}(y) = \cos x \cdot 1 = \cos x$

(3) $z = \frac{1}{\sqrt{x^2 + y^2}} = (x^2 + y^2)^{-1/2}$

• 对 $x$ 求偏导：
  $z_x = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot \frac{\partial}{\partial x}(x^2 + y^2) = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot 2x = -x(x^2 + y^2)^{-3/2} = \frac{-x}{(x^2 + y^2)^{3/2}}$

• 对 $y$ 求偏导：
  $z_y = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot \frac{\partial}{\partial y}(x^2 + y^2) = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot 2y = -y(x^2 + y^2)^{-3/2} = \frac{-y}{(x^2 + y^2)^{3/2}}$

(4) $z = \ln(x + y^2)$

• 对 $x$ 求偏导：
  $z_x = \frac{1}{x + y^2} \cdot \frac{\partial}{\partial x}(x + y^2) = \frac{1}{x + y^2} \cdot 1 = \frac{1}{x + y^2}$

• 对 $y$ 求偏导：
  $z_y = \frac{1}{x + y^2} \cdot \frac{\partial}{\partial y}(x + y^2) = \frac{1}{x + y^2} \cdot 2y = \frac{2y}{x + y^2}$

(5) $z = e^{xy}$

• 对 $x$ 求偏导：
  $z_x = e^{xy} \cdot \frac{\partial}{\partial x}(xy) = e^{xy} \cdot y = y e^{xy}$

• 对 $y$ 求偏导：
  $z_y = e^{xy} \cdot \frac{\partial}{\partial y}(xy) = e^{xy} \cdot x = x e^{xy}$

(6) $z = \arctan \frac{y}{x}$

• 对 $x$ 求偏导：
  利用公式 $(\arctan u)' = \frac{1}{1 + u^2} \cdot u'$，其中 $u = \frac{y}{x}$。
  $z_x = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) = \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2 + y^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}$

• 对 $y$ 求偏导：
  $z_y = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial y}\left(\frac{y}{x}\right) = \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}$