Question

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

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

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

Explanation:

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

To find the first partial derivative with respect to \( x \), we use the product rule. The product rule states that if \( u(x,y) = 2x \) and \( v(x,y)=e^{xy} \), then \( (uv)'_x = u'_x v + u v'_x \).

The derivative of \( u = 2x \) with respect to \( x \) is \( u'_x = 2 \). For \( v = e^{xy} \), using the chain rule, the derivative with respect to \( x \) is \( v'_x = y e^{xy} \) (since the derivative of \( e^{u} \) with respect to \( x \) is \( e^{u} \cdot u'_x \), here \( u = xy \), so \( u'_x = y \)).

So, \( f_x(x,y) = 2 \cdot e^{xy} + 2x \cdot y e^{xy} = 2e^{xy} + 2xy e^{xy} \).

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

Now we take the partial derivative of \( f_x(x,y) \) with respect to \( x \). Let's denote \( f_x = 2e^{xy} + 2xy e^{xy} \). We will differentiate each term separately.

First term: derivative of \( 2e^{xy} \) with respect to \( x \). Using the chain rule, the derivative of \( e^{xy} \) with respect to \( x \) is \( y e^{xy} \), so the derivative of \( 2e^{xy} \) is \( 2y e^{xy} \).

Second term: derivative of \( 2xy e^{xy} \) with respect to \( x \). We use the product rule here. Let \( u = 2xy \) and \( v = e^{xy} \). Then \( u'_x = 2y \) and \( v'_x = y e^{xy} \). By the product rule \( (uv)'_x = u'_x v + u v'_x \), we have:

\( (2xy e^{xy})'_x = 2y \cdot e^{xy} + 2xy \cdot y e^{xy} = 2y e^{xy} + 2x y^2 e^{xy} \).

Now, add the derivatives of the two terms together:

\( f_{xx}(x,y) = 2y e^{xy} + 2y e^{xy} + 2x y^2 e^{xy} \).

Combine like terms: \( 2y e^{xy} + 2y e^{xy} = 4y e^{xy} \), so:

\( f_{xx}(x,y) = 4y e^{xy} + 2x y^2 e^{xy} \). We can factor out \( 2y e^{xy} \):

\( f_{xx}(x,y) = 2y e^{xy}(2 + x y) \) or we can write it as \( 2y e^{xy} + 2xy^2 e^{xy} \) (or factor further, but this is the simplified form).

Answer:

\( 2y e^{xy} + 2xy^2 e^{xy} \) (or \( 2y e^{xy}(2 + xy) \))