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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$f_{xx}(x,y) = \square$

Explanation:

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

To find the first - order partial derivative with respect to \( x \), we treat \( y \) as a constant.
For the term \( 3x^{2}y^{2} \), using the power rule \( \frac{\partial}{\partial x}(ax^{n})=nax^{n - 1} \), we have \( \frac{\partial}{\partial x}(3x^{2}y^{2})=3\times2x^{2 - 1}y^{2}=6xy^{2} \).
For the term \( 5x^{4} \), \( \frac{\partial}{\partial x}(5x^{4}) = 5\times4x^{4-1}=20x^{3} \).
For the term \( 4y \), since \( y \) is treated as a constant, \( \frac{\partial}{\partial x}(4y) = 0 \).
So, \( f_x(x,y)=6xy^{2}+20x^{3} \).

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

Now, we take the partial derivative of \( f_x(x,y) \) with respect to \( x \).
For the term \( 6xy^{2} \), treating \( y \) as a constant, \( \frac{\partial}{\partial x}(6xy^{2})=6y^{2} \).
For the term \( 20x^{3} \), using the power rule, \( \frac{\partial}{\partial x}(20x^{3})=20\times3x^{3 - 1}=60x^{2} \).
So, \( f_{xx}(x,y)=6y^{2}+60x^{2} \).

Answer:

\( 6y^{2}+60x^{2} \)