4.2 Partial derivatives

1.

(a) $${z_{x} = 3x^2 + y^2, z_{y} = 2xy + 3y^2}$$

(b) $${z_{x} = e^{x}\sin{y}, z_{y} = e^{x}\cos{y}}$$

(c) $${z_{x} = \frac{2x}{x^2 + y^2}, z_{y} = x^3 + 2xy}$$

2.

(a) $${z_{x} = 3x^2 y + y^2, z_{y} = x^3 + 2xy, z_{xx} = 6xy, z_{xy} = 3x^2 + 2y, z_{yx} = 3x^2 + 2y, z_{yy} = 2x}$$

(b) $${(y^2 + xy)e^{\frac{x}{y}}, z_{y} = (2xy - x^2)e^{\frac{x}{y}}, z_{xx} = e^{\frac{x}{y}}(x + 2y)}$$

$${z_{xy} = e^{\frac{x}{y}}(2y - \frac{x^2}{y}), z_{yy} = e^{\frac{x}{y}}(2x - \frac{2x^2}{y} + \frac{x^3}{y^2})}$$

(c) $${z_{x} = \frac{2x}{1 + (x^2 + y^2)^2}, z_{y} = \frac{2y}{1 + (x^... ... z_{xx} = \frac{2 + 2(x^2 + y^2)^2 - 8x^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}}$$

$${z_{yy} = \frac{2 + 2(x^2 + y^2)^2 - 8y^2(x^2 + y^2)}{(1 + (x^2 + y^2)^2)}, z_{xy} = \frac{ - 8x(x^2 + y^2)}{(1 + (x^2 + y^2)^2)^2}}$$

3.

(a) Check to see $f_{x}(0,0), f_{y}(0,0)$ exist． $f{x}(x,0) = \frac{0}{x^2} = 0$ implies $f_{x}(x,0) = 0$．Thus， $f_{x}(0,0) = 0$． $f{y}(0,y) = \frac{y^3}{y^2} = y$． $f_{y}(0,y) = 1$．Therefore， $f_{y}(0,0) = 1$．

(b) $f(x,0) = \log{1} = 0$ implies $f_{x}(x,0) = 0$. Then， $f_{x}(0,0) = 0$. $f(0,y) = \log(1+y^2)$ implies $f_{y}(0,y) = \frac{2y}{1 + y^2}$．Thus， $f_{y}(0,0) = 0$.