Answer
$$dz = \left( {10x{y^5} - 2} \right)dx + \left( {25{x^2}{y^4} + 4} \right)dy$$
Work Step by Step
$$\eqalign{
& z = 5{x^2}{y^5} - 2x + 4y + 7 \cr
& {\text{Let }}z = f\left( {x,y} \right):{\text{ }} \cr
& {\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right){\text{, treating }}y{\text{ as a constant}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^2}{y^5} - 2x + 4y + 7} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = 5{y^5}\left( {2x} \right) - 2 \cr
& {\text{ }}{f_x}\left( {x,y} \right) = 10x{y^5} - 2 \cr
& {\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right){\text{, treating }}x{\text{ as a constant}} \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^2}{y^5} - 2x + 4y + 7} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = 5{x^2}\left( {5{y^4}} \right) + 4 \cr
& {\text{ }}{f_y}\left( {x,y} \right) = 25{x^2}{y^4} + 4 \cr
& \cr
& {\text{The total differential of }}z{\text{ is given by }}dz = {f_x}\left( {x,y} \right)dx + {f_y}\left( {x,y} \right)dy \cr
& {\text{Substitute the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& dz = \left( {10x{y^5} - 2} \right)dx + \left( {25{x^2}{y^4} + 4} \right)dy \cr} $$
