Question

find the second - order partial derivative.
find ( f_{xy} ) when ( f(x,y)=10x^{2}y^{4}-7x^{3}y^{5} ).
a. ( 80xy^{3}-105x^{2}y^{4} )
b. ( 160xy^{3}-105x^{2}y^{4} )
c. ( 80xy^{3}-21x^{2}y^{4} )
d. ( 160xy^{3}-21x^{2}y^{4} )

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