Question

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.)

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:

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