Question

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy

Answer 1

ANSWER TO QUESTION 1

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

$f_{x} = 2y {e}^{xy}$

and the second derivative with respect to x is,

$f_{xx} = 2 {y}^{2} {e}^{xy}$

ANSWER TO QUESTION 2

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

$f_{y} = 2x{e}^{xy}$

and the second derivative with respect to y is

$f_{yy} = 2 {x}^{2} {e}^{xy}$

ANSWER TO QUESTION 3

Our first mixed partial is

$f_{xy}$

We need to differentiate
$f_{x} = 2y {e}^{xy}$
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,

$f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'$

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}$

ANSWER TO QUESTION 4

The second mixed partial is

$f_{yx}$

We need to differentiate
$f_{y} = 2x{e}^{xy}$

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,


$f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'$

$f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}$

Hence,

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}$