Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 39

$${f_{xx}}\left( {x,y} \right) = \frac{{12y}}{{{{\left( {2x - y} \right)}^3}}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = \frac{{ - 6x - 3y}}{{{{\left( {2x - y} \right)}^3}}}$$

$$\eqalign{ & f\left( {x,y} \right) = \frac{{3x}}{{2x - y}} \cr & {\text{find }}{f_x}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{3x}}{{2x - y}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then by using the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {2x - y} \right)\frac{\partial }{{\partial x}}\left[ {3x} \right] - 3x\frac{\partial }{{\partial x}}\left[ {2x - y} \right]}}{{{{\left( {2x - y} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {2x - y} \right)\left( 3 \right) - 3x\left( 2 \right)}}{{{{\left( {2x - y} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_x}\left( {x,y} \right) = \frac{{6x - 3y - 6x}}{{{{\left( {2x - y} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{ - 3y}}{{{{\left( {2x - y} \right)}^2}}} \cr & \cr & {\text{find }}{f_{xx}}\left( {x,y} \right) \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{ - 3y}}{{{{\left( {2x - y} \right)}^2}}}} \right] \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 3y{{\left( {2x - y} \right)}^{ - 2}}} \right] \cr & {\text{by using the chain rule}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 6y{\left( {2x - y} \right)^{ - 3}}\frac{\partial }{{\partial x}}\left[ {2x - y} \right] \cr & {\text{solving derivatives}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 6y{\left( {2x - y} \right)^{ - 3}}\left( 2 \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{{12y}}{{{{\left( {2x - y} \right)}^3}}} \cr & \cr & {\text{find }}{f_{xy}}\left( {x,y} \right) \cr & {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{ - 3y}}{{{{\left( {2x - y} \right)}^2}}}} \right] \cr & {\text{by using the quotient rule}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{{{\left( {2x - y} \right)}^2}\left( { - 3} \right) + 3y\left( 2 \right)\left( {2x - y} \right)\left( { - 1} \right)}}{{{{\left( {2x - y} \right)}^4}}} \cr & {\text{simplifying}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{\left( {2x - y} \right)\left( { - 3} \right) - 6y}}{{{{\left( {2x - y} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{ - 6x + 3y - 6y}}{{{{\left( {2x - y} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{ - 6x - 3y}}{{{{\left( {2x - y} \right)}^3}}} \cr} $$