Question

find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(2,1)$ and $f_y(-3,-4)$. $f(x,y)=2x^2y^3$ $f_x(x,y)=\square$

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:

\( f_x(x,y)=4xy^{3} \), \( f_y(x,y)=6x^{2}y^{2} \), \( f_x(2,1)=8 \), \( f_y(-3,-4)=864 \)