Answer
$$\eqalign{
& f\left( {x,y} \right) = \sin \left( {x + y} \right) \cr
& dz = 0 \cr} $$
Work Step by Step
$$\eqalign{
& \sin \left[ {{{\left( {1.05} \right)}^2} + {{\left( {0.95} \right)}^2}} \right] - \sin \left( {{1^2} + {1^2}} \right) \cr
& {\text{Rewrite the expression}} \cr
& = \sin \left[ {{{\left( {1 + 0.05} \right)}^2} + {{\left( {1 - 0.05} \right)}^2}} \right] - \sin \left( {{1^2} + {1^2}} \right) \cr
& = \sin \left[ {\overbrace {{{\left( {1 + 0.05} \right)}^2}}^{{{\left( {x + \Delta x} \right)}^2}} + \overbrace {{{\left( {1 - 0.05} \right)}^2}}^{{{\left( {y + \Delta y} \right)}^2}}} \right] - \overbrace {\sin \left( {{1^2} + {1^2}} \right)}^{\sin \left( {x + y} \right)} \cr
& {\text{Let the function }} \cr
& z = f\left( {x,y} \right) = \sin \left( {x + y} \right),{\text{ with }}x = 1{\text{ and }}y = 1,{\text{ }} \cr
& \Delta x = 0.05{\text{ and }}\Delta y = - 0.05 \cr
& {\text{Therefore, the total differential is }} \cr
& dz = \frac{\partial }{{\partial x}}\left[ {\sin \left( {x + y} \right)} \right]dx + \frac{\partial }{{\partial y}}\left[ {\sin \left( {x + y} \right)} \right]dy \cr
& dz = \cos \left( {x + y} \right)dx + \cos \left( {x + y} \right)dy \cr
& {\text{With }}dx = \Delta x{\text{ and }}dy = \Delta y \cr
& dz = \cos \left( {1 + 1} \right)\left( {0.05} \right) + \cos \left( {1 + 1} \right)\left( { - 0.05} \right) \cr
& dz = 0 \cr} $$
