Answer
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr
& dz = 0.943 \cr} $$
Work Step by Step
$$\eqalign{
& \sqrt {{{\left( {5.05} \right)}^2} + {{\left( {3.1} \right)}^2}} - \sqrt {{5^2} + {3^2}} \cr
& {\text{Rewrite the expression}} \cr
& = \sqrt {{{\left( {5 + 0.05} \right)}^2} + {{\left( {3 + 0.1} \right)}^2}} - \sqrt {{5^2} + {3^2}} \cr
& = \sqrt {\underbrace {{{\left( {5 + 0.05} \right)}^2}}_{{{\left( {x + \Delta x} \right)}^2}} + \underbrace {{{\left( {3 + 0.1} \right)}^2}}_{{{\left( {y + \Delta y} \right)}^2}}} - \underbrace {\sqrt {{5^2} + {3^2}} }_{\sqrt {{x^2} + {y^2}} } \cr
& {\text{Let the function }} \cr
& z = f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} ,{\text{ with }}x = 5{\text{ and }}y = 3,{\text{ }} \cr
& \Delta x = 0.05{\text{ and }}\Delta y = 0.1 \cr
& {\text{Therefore, the total differential is }} \cr
& dz = \frac{\partial }{{\partial x}}\left[ {\sqrt {{x^2} + {y^2}} } \right]dx + \frac{\partial }{{\partial y}}\left[ {\sqrt {{x^2} + {y^2}} } \right]dy \cr
& dz = \frac{x}{{\sqrt {{x^2} + {y^2}} }}dx + \frac{y}{{\sqrt {{x^2} + {y^2}} }}dy \cr
& {\text{With }}dx = \Delta x{\text{ and }}dy = \Delta y \cr
& dz = \frac{5}{{\sqrt {{5^2} + {3^2}} }}\left( {0.05} \right) + \frac{3}{{\sqrt {{5^2} + {3^2}} }}\left( {0.1} \right) \cr
& dz = 0.0943 \cr
& \cr
& f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr
& dz = 0.943 \cr} $$
