Answer
$${f_x}\left( {x,y} \right) = 1{\text{ and }}{f_y}\left( {x,y} \right) = 2y$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = x + {y^2} + 4 \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h} \right) + {y^2} + 4 - \left( {x + {y^2} + 4} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{x + h + {y^2} + 4 - x - {y^2} - 4}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} 1 \cr
& {\text{Evaluate the limit}} \cr
& {f_x}\left( {x,y} \right) = 1 \cr
& \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x,y + h} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{x + {{\left( {y + h} \right)}^2} + 4 - \left( {x + {y^2} + 4} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{x + {y^2} + 2yh + {h^2} + 4 - x - {y^2} - 4}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2yh + {h^2}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {2y + h} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \left( {2y + h} \right) \cr
& {\text{Evaluate the limit}} \cr
& {f_y}\left( {x,y} \right) = 2y \cr} $$
