(a) A aproximação de primeira ordem de
f em
(x, y)=(2,-2)
\begin{aligned} \frac{\partial f}{\partial x}(x, y) & =\frac{\partial}{\partial x}\left(\frac{x+y}{x-y}\right) \\\ & =\frac{\frac{\partial}{\partial x}(x+y)(x-y)-(x+y) \frac{\partial}{\partial x}(x-y)}{(x-y)^{2}} \\\ & =\frac{(1)(x-y)-(x+y)(1)}{(x-y)^{2}} \\\ & =\frac{-2 y}{(x-y)^{2}} \\\ \frac{\partial f}{\partial y}(x, y) & =\frac{\partial}{\partial y}\left(\frac{x+y}{x-y}\right) \\\ & =\frac{\frac{\partial}{\partial y}(x+y)(x-y)-(x+y) \frac{\partial}{\partial y}(x-y)}{(x-y)^{2}} \\\ & =\frac{(1)(x-y)-(x+y)(-1)}{(x-y)^{2}} \\\ & =\frac{2 x}{(x-y)^{2}} \end{aligned}
Portanto
\nabla f(x, y)=\left(\frac{-2 y}{(x-y)^{2}}, \frac{2 x}{(x-y)^{2}}\right).
\begin{aligned} f(2,-2) & =\frac{2+(-2)}{2-(-2)}=0 \\\ \nabla f(2,-2) & =\left(\frac{-2 \cdot(-2)}{(2-(-2))^{2}}, \frac{2 \cdot 2}{(2-(-2))^{2}}\right)=\left(\frac{1}{4}, \frac{1}{4}\right) \end{aligned}
Segue que a aproximação de primeira ordem em
(2,-2) é:
\begin{aligned} g(x, y) & =\langle\nabla f(2,-2),(x-2, y-(-2))\rangle+f(2,-2) \\\ & =\left\langle\left(\frac{1}{4}, \frac{1}{4}\right),(x-2, y+2)\right\rangle+0 \\\ g(x, y) & =\frac{1}{4}(x-2)+\frac{1}{4}(y+2) \end{aligned}