Answer
The solutions are:
a)
$$\frac{f(x+\Delta x,y)- f(x,y)}{\Delta x} =2.$$
b)
$$\frac{f(x,y+\Delta y)- f(x,y)}{\Delta y} = 2y+\Delta y.$$
Work Step by Step
a) We have that
$$f(x+\Delta x,y) = 2(x+\Delta x)+y^2$$
and this gives
$$\frac{f(x+\Delta x,y)- f(x,y)}{\Delta x} = \frac{2(x+\Delta x)+y^2-(2x+y^2)}{\Delta x} = \frac{2x+2\Delta x +y^2-2x-y^2}{\Delta x} =\frac{2\Delta x}{\Delta x} = 2.$$
b) We have that
$$f(x,y+\Delta y)- f(x,y) = 2x+(y+\Delta y)^2 = 2x + y^2+2y\Delta y + \Delta y^2 $$
and this gives
$$\frac{f(x,y+\Delta y)- f(x,y)}{\Delta y} = \frac{2x + y^2+2y\Delta y + \Delta y^2
- (2x+y^2)}{\Delta y} =\frac{2x + y^2+2y\Delta y + \Delta y^2
- 2x-y^2}{\Delta y}=\frac{2y\Delta y+\Delta y^2}{\Delta y} = 2y+\Delta y.$$
