Question

find $f_{xx}(x,y)$, $f_{xy}(x,y)$, $f_{yx}(x,y)$, and $f_{yy}(x,y)$ for the following function.\
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$f(x,y) = 3x^2y^2 + 5x^4 + 4y$\
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$f_{xx}(x,y) = 6y^2 + 60x^2$\
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$f_{xy}(x,y) = 12xy$\
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$f_{yx}(x,y) = 12xy$\
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$f_{yy}(x,y) = \square$

Explanation:

Step1: Find \( f_y(x,y) \)

First, find the first - order partial derivative with respect to \( y \). For the function \( f(x,y)=3x^{2}y^{2}+5x^{4}+4y \), when we take the partial derivative with respect to \( y \), we treat \( x \) as a constant.
Using the power rule \( \frac{\partial}{\partial y}(y^{n}) = ny^{n - 1} \) and the sum rule of differentiation:
\( f_y(x,y)=\frac{\partial}{\partial y}(3x^{2}y^{2})+\frac{\partial}{\partial y}(5x^{4})+\frac{\partial}{\partial y}(4y) \)
\( = 3x^{2}\times2y+0 + 4=6x^{2}y + 4 \)

Step2: Find \( f_{yy}(x,y) \)

Now, take the partial derivative of \( f_y(x,y) = 6x^{2}y+4 \) with respect to \( y \). Treat \( x \) as a constant.
\( f_{yy}(x,y)=\frac{\partial}{\partial y}(6x^{2}y + 4)=\frac{\partial}{\partial y}(6x^{2}y)+\frac{\partial}{\partial y}(4) \)
\( = 6x^{2}\times1+0 = 6x^{2} \)

Answer:

\( f_{yy}(x,y)=6x^{2} \)