find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(2,1)$ and $f_y(-3,-4)$. $f(x,y)=2x^2y^3$ $f_x(x,y)=\\square$
find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(2,1)$ and $f_y(-3,-4)$. $f(x,y)=2x^2y^3$ $f_x(x,y)=\\square$
Answer
Explanation:
Step1: Find ( f_x(x,y) )
Differentiate ( f(x,y) = 2x^{2}y^{3} ) with respect to ( x ) (treating ( y ) as a constant). Using the power rule ( \frac{d}{dx}(ax^{n})=nax^{n - 1} ), we have ( f_x(x,y)=\frac{\partial}{\partial x}(2x^{2}y^{3})=2y^{3}\frac{\partial}{\partial x}(x^{2}) ). Since ( \frac{\partial}{\partial x}(x^{2}) = 2x ), then ( f_x(x,y)=4xy^{3} ).
Step2: Find ( f_y(x,y) )
Differentiate ( f(x,y) = 2x^{2}y^{3} ) with respect to ( y ) (treating ( x ) as a constant). Using the power rule ( \frac{d}{dy}(ay^{n})=nay^{n - 1} ), we have ( f_y(x,y)=\frac{\partial}{\partial y}(2x^{2}y^{3})=2x^{2}\frac{\partial}{\partial y}(y^{3}) ). Since ( \frac{\partial}{\partial y}(y^{3}) = 3y^{2} ), then ( f_y(x,y)=6x^{2}y^{2} ).
Step3: Find ( f_x(2,1) )
Substitute ( x = 2 ) and ( y = 1 ) into ( f_x(x,y)=4xy^{3} ). ( f_x(2,1)=4\times2\times1^{3}=8 ).
Step4: Find ( f_y(-3,-4) )
Substitute ( x=-3 ) and ( y = - 4 ) into ( f_y(x,y)=6x^{2}y^{2} ). First, ( x^{2}=(-3)^{2}=9 ) and ( y^{2}=(-4)^{2}=16 ). Then ( f_y(-3,-4)=6\times9\times16=864 ).
Answer:
( f_x(x,y)=4xy^{3} ), ( f_y(x,y)=6x^{2}y^{2} ), ( f_x(2,1)=8 ), ( f_y(-3,-4)=864 )