find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(-2,2)$ and $f_y(2,-3)$. $f(x,y)=5xy + 4y^4+4$…

find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(-2,2)$ and $f_y(2,-3)$. $f(x,y)=5xy + 4y^4+4$ $f_x(x,y)=\\square$ $f_y(x,y)=\\square$ $f_x(-2,2)=\\square$ (simplify your answer.) $f_y(2,-3)=\\square$ (simplify your answer.)

find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(-2,2)$ and $f_y(2,-3)$. $f(x,y)=5xy + 4y^4+4$ $f_x(x,y)=\\square$ $f_y(x,y)=\\square$ $f_x(-2,2)=\\square$ (simplify your answer.) $f_y(2,-3)=\\square$ (simplify your answer.)

Answer

Explanation:

Step1: Find ( f_x(x,y) )

Differentiate ( f(x,y) = 5xy+4y^{4}+4 ) with respect to ( x ). Using the rule ( \frac{\partial}{\partial x}(ax^n y^m)=a n x^{n - 1}y^m ) (where ( n = 1,m = 1) for the (5xy) term and (y) is treated as a constant), ( \frac{\partial}{\partial x}(4y^{4})=0) (since derivative of a constant with respect to (x) is (0)) and ( \frac{\partial}{\partial x}(4)=0). ( f_x(x,y)=\frac{\partial}{\partial x}(5xy)+\frac{\partial}{\partial x}(4y^{4})+\frac{\partial}{\partial x}(4)=5y)

Step2: Find ( f_y(x,y) )

Differentiate ( f(x,y) = 5xy+4y^{4}+4 ) with respect to ( y ). Using the rule ( \frac{\partial}{\partial y}(ax^n y^m)=a m x^{n}y^{m - 1}), ( \frac{\partial}{\partial y}(5xy)=5x), ( \frac{\partial}{\partial y}(4y^{4})=16y^{3}), ( \frac{\partial}{\partial y}(4)=0). ( f_y(x,y)=\frac{\partial}{\partial y}(5xy)+\frac{\partial}{\partial y}(4y^{4})+\frac{\partial}{\partial y}(4)=5x + 16y^{3})

Step3: Find ( f_x(-2,2) )

Substitute ( x=-2,y = 2) into ( f_x(x,y)=5y). ( f_x(-2,2)=5\times2 = 10)

Step4: Find ( f_y(2,-3) )

Substitute ( x = 2,y=-3) into ( f_y(x,y)=5x + 16y^{3}). ( f_y(2,-3)=5\times2+16\times(-3)^{3}=10+16\times(-27)=10-432=-422)

Answer:

( f_x(x,y)=5y), ( f_y(x,y)=5x + 16y^{3}), ( f_x(-2,2)=10), ( f_y(2,-3)=-422)