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

Explanation:

Step1: Recall Clairaut's Theorem

By Clairaut's Theorem, for a function \( f(x,y) \) with continuous second - order partial derivatives, \( f_{xy}(x,y)=f_{yx}(x,y) \).

Step2: Use the given value of \( f_{xy}(x,y) \)

We are given that \( f_{xy}(x,y) = 12xy \). Since \( f_{xy} \) and \( f_{yx} \) are equal (by Clairaut's Theorem, and we can also verify by computing \( f_{yx} \) directly:

First, find \( f_y(x,y) \):
Given \( f(x,y)=3x^{2}y^{2}+5x^{4}+4y \), we differentiate with respect to \( y \):
\( f_y(x,y)=\frac{\partial}{\partial y}(3x^{2}y^{2}+5x^{4}+4y)=3x^{2}\times2y + 0+4 = 6x^{2}y + 4 \)

Then, differentiate \( f_y(x,y) \) with respect to \( x \):
\( f_{yx}(x,y)=\frac{\partial}{\partial x}(6x^{2}y + 4)=6\times2xy+0 = 12xy \)

Answer:

\( f_{yx}(x,y)=12xy \)