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Find-Z-x-and-Z-y-for-each-of-the-functions-below-a-Z-8x-2-y-14xy-2-5y-2-x-3-b-Z-4x-3-y-2-2x-2-y-3-7xy-5-




Question Number 18469 by tawa tawa last updated on 22/Jul/17
Find Z_x  and Z_y  for each of the functions below  (a)  Z = 8x^2 y + 14xy^2  + 5y^2 x^3   (b)  Z = 4x^3 y^2  + 2x^2 y^3  − 7xy^5
$$\mathrm{Find}\:\mathrm{Z}_{\mathrm{x}} \:\mathrm{and}\:\mathrm{Z}_{\mathrm{y}} \:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{below} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{Z}\:=\:\mathrm{8x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{14xy}^{\mathrm{2}} \:+\:\mathrm{5y}^{\mathrm{2}} \mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{Z}\:=\:\mathrm{4x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \:−\:\mathrm{7xy}^{\mathrm{5}} \\ $$
Answered by Tinkutara last updated on 22/Jul/17
(a) Z_x  = 8y(2x) + 14y^2  + 5y^2 (3x^2 )  = 16xy + 14y^2  + 15x^2 y^2   Z_y  = 8x^2  + 14x(2y) + 5x^3 (2y)  = 10x^3 y + 8x^2  + 28xy  (b) Z_x  = 4y^2 (3x^2 ) + 2y^3 (2x) − 7y^5   = 12x^2 y^2  − 7y^5  + 4xy^3   Z_y  = 4x^3 (2y) + 2x^2 (3y^2 ) − 7x(5y^4 )  = −35xy^4  + 6x^2 y^2  + 8x^3 y
$$\left(\mathrm{a}\right)\:{Z}_{{x}} \:=\:\mathrm{8}{y}\left(\mathrm{2}{x}\right)\:+\:\mathrm{14}{y}^{\mathrm{2}} \:+\:\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} \right) \\ $$$$=\:\mathrm{16}{xy}\:+\:\mathrm{14}{y}^{\mathrm{2}} \:+\:\mathrm{15}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$${Z}_{{y}} \:=\:\mathrm{8}{x}^{\mathrm{2}} \:+\:\mathrm{14}{x}\left(\mathrm{2}{y}\right)\:+\:\mathrm{5}{x}^{\mathrm{3}} \left(\mathrm{2}{y}\right) \\ $$$$=\:\mathrm{10}{x}^{\mathrm{3}} {y}\:+\:\mathrm{8}{x}^{\mathrm{2}} \:+\:\mathrm{28}{xy} \\ $$$$\left(\mathrm{b}\right)\:{Z}_{{x}} \:=\:\mathrm{4}{y}^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} \right)\:+\:\mathrm{2}{y}^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:−\:\mathrm{7}{y}^{\mathrm{5}} \\ $$$$=\:\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:−\:\mathrm{7}{y}^{\mathrm{5}} \:+\:\mathrm{4}{xy}^{\mathrm{3}} \\ $$$${Z}_{{y}} \:=\:\mathrm{4}{x}^{\mathrm{3}} \left(\mathrm{2}{y}\right)\:+\:\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{3}{y}^{\mathrm{2}} \right)\:−\:\mathrm{7}{x}\left(\mathrm{5}{y}^{\mathrm{4}} \right) \\ $$$$=\:−\mathrm{35}{xy}^{\mathrm{4}} \:+\:\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:\mathrm{8}{x}^{\mathrm{3}} {y} \\ $$
Commented by tawa tawa last updated on 22/Jul/17
i really appreciate sir. God bless you
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

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