find the second - order partial derivative.\nfind ( f_{xy} ) when ( f(x,y)=10x^{2}y^{4}-7x^{3}y^{5} ).\na. (…

find the second - order partial derivative.\nfind ( f_{xy} ) when ( f(x,y)=10x^{2}y^{4}-7x^{3}y^{5} ).\na. ( 80xy^{3}-105x^{2}y^{4} )\nb. ( 160xy^{3}-105x^{2}y^{4} )\nc. ( 80xy^{3}-21x^{2}y^{4} )\nd. ( 160xy^{3}-21x^{2}y^{4} )

find the second - order partial derivative.\nfind ( f_{xy} ) when ( f(x,y)=10x^{2}y^{4}-7x^{3}y^{5} ).\na. ( 80xy^{3}-105x^{2}y^{4} )\nb. ( 160xy^{3}-105x^{2}y^{4} )\nc. ( 80xy^{3}-21x^{2}y^{4} )\nd. ( 160xy^{3}-21x^{2}y^{4} )

Answer

Explanation:

Step1: Find the first - order partial derivative (f_x)

Differentiate (f(x,y)=10x^{2}y^{4}-7x^{3}y^{5}) with respect to (x) (treating (y) as a constant). Using the power rule (\frac{\partial}{\partial x}(ax^{n}y^{m}) = anx^{n - 1}y^{m}), we have: (f_x=\frac{\partial}{\partial x}(10x^{2}y^{4}-7x^{3}y^{5})=10\times2x^{1}y^{4}-7\times3x^{2}y^{5}=20xy^{4}-21x^{2}y^{5})

Step2: Find the second - order partial derivative (f_{xy})

Differentiate (f_x = 20xy^{4}-21x^{2}y^{5}) with respect to (y) (treating (x) as a constant). Using the power rule (\frac{\partial}{\partial y}(ax^{n}y^{m})=amx^{n}y^{m - 1}), we get: (f_{xy}=\frac{\partial}{\partial y}(20xy^{4}-21x^{2}y^{5})=20x\times4y^{3}-21x^{2}\times5y^{4}=80xy^{3}-105x^{2}y^{4})

Answer:

A. (80xy^{3}-105x^{2}y^{4})