Question

find $f_{xx}(x,y)$, $f_{xy}(x,y)$, $f_{yx}(x,y)$, and $f_{yy}(x,y)$ for the following function.

$f(x,y) = 3x^2y^2 + 5x^4 + 4y$

$f_{xx}(x,y) = 6y^2 + 60x^2$

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

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 function \( f(x,y)=3x^{2}y^{2}+5x^{4}+4y \), using the power rule \( \frac{d}{dx}(x^{n}) = nx^{n - 1} \):

• The derivative of \( 3x^{2}y^{2} \) with respect to \( x \) is \( 3\times2x^{2 - 1}y^{2}=6xy^{2} \) (since \( y^{2} \) is treated as a constant).
• The derivative of \( 5x^{4} \) with respect to \( x \) is \( 5\times4x^{4 - 1}=20x^{3} \).
• The derivative of \( 4y \) with respect to \( x \) is \( 0 \) (because \( 4y \) does not contain \( x \)).

So, \( f_x(x,y)=6xy^{2}+20x^{3} \).

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

Now, we take the partial derivative of \( f_x(x,y) = 6xy^{2}+20x^{3} \) with respect to \( y \). We treat \( x \) as a constant.

• The derivative of \( 6xy^{2} \) with respect to \( y \) is \( 6x\times2y^{2 - 1}=12xy \) (since \( x \) is treated as a constant).
• The derivative of \( 20x^{3} \) with respect to \( y \) is \( 0 \) (because \( 20x^{3} \) does not contain \( y \)).

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

Answer:

\( 12xy \)