ANSWER TO QUESTION 1
The given function is
[tex]f(x,y)=2 {e}^{xy} [/tex]
The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is
[tex]f_{x} = 2y {e}^{xy} [/tex]
and the second derivative with respect to x is,
[tex]f_{xx} = 2 {y}^{2} {e}^{xy} [/tex]
ANSWER TO QUESTION 2
The given function is
[tex]f(x,y)=2 {e}^{xy} [/tex]
The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is
[tex]f_{y} = 2x{e}^{xy} [/tex]
and the second derivative with respect to y is
[tex]f_{yy} = 2 {x}^{2} {e}^{xy} [/tex]
ANSWER TO QUESTION 3
Our first mixed partial is
[tex]f_{xy} [/tex]
We need to differentiate
[tex]f_{x} = 2y {e}^{xy} [/tex]
again. But this time with respect to y.
Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,
[tex]f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'[/tex]
[tex]f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}[/tex]
ANSWER TO QUESTION 4
The second mixed partial is
[tex]f_{yx} [/tex]
We need to differentiate
[tex]f_{y} = 2x{e}^{xy} [/tex]
again. But this time with respect to x.
Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,
[tex]f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'[/tex]
[tex]f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}[/tex]
Hence,
[tex]f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}[/tex]
