Answer
$$dz = 20{x^3}{y^3}dx + 15{x^4}{y^2}dy$$
Work Step by Step
$$\eqalign{
& z = 5{x^4}{y^3} \cr
& {\text{The total differential of the dependent variable }}z{\text{ is}} \cr
& dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy \cr
& {\text{Calculating }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^3}} \right] \cr
& \frac{{\partial z}}{{\partial x}} = 20{x^3}{y^3} \cr
& and \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^3}} \right] \cr
& \frac{{\partial z}}{{\partial y}} = 15{x^4}{y^2} \cr
& {\text{Therefore,}} \cr
& dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy \cr
& dz = 20{x^3}{y^3}dx + 15{x^4}{y^2}dy \cr} $$
