To determine whether the given derivatives are correct, we proceed step by step. The function provided is:

$f(x,y) = xye^{x^2}.$

We need to verify the following:

1. $f_x = xy(2x \cdot e^{x^2}) + e^{x^2}(y)$
2. $f_{xy} = ye^{x^2}(2x^2 + 1)$.

Step 1: Compute $f_x$ (the partial derivative with respect to $x$)

The function $f(x,y) = xye^{x^2}$ can be differentiated with respect to $x$. Here, $y$ is treated as a constant. We use the product rule:

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

Let $u = xy$ and $v = e^{x^2}$. The derivative of a product is:

$\frac{\partial}{\partial x}[uv] = u'v + uv'.$

• $u = xy \implies u' = y$ (since $y$ is constant).
• $v = e^{x^2} \implies v' = e^{x^2} \cdot 2x$ (by the chain rule).

Thus:

$f_x = (u'v) + (uv') = (y \cdot e^{x^2}) + (xy \cdot 2x e^{x^2}).$

Simplify:

$f_x = y e^{x^2} + 2x^2 y e^{x^2}.$

Factor out $y e^{x^2}$:

$f_x = y e^{x^2} (1 + 2x^2).$

Step 2: Compare with the given $f_x$

The problem states:

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

Simplify the given form:

• The term $xy(2x \cdot e^{x^2})$ simplifies to $2x^2 y e^{x^2}$,
• The term $e^{x^2}(y)$ simplifies to $y e^{x^2}$.

Combining these:

$f_x = y e^{x^2} + 2x^2 y e^{x^2}.$

This matches the expression we derived for $f_x$. Therefore, the given $f_x$ is correct.

Step 3: Compute $f_{xy}$ (the mixed second derivative)

Now we differentiate $f_x = y e^{x^2} (1 + 2x^2)$ with respect to $y$. Recall:

• $y e^{x^2} (1 + 2x^2)$ is a product, and we treat $x$ as a constant when differentiating with respect to $y$.

Only the $y$-term depends on $y$, so:

$f_{xy} = \frac{\partial}{\partial y} \left[ y e^{x^2} (1 + 2x^2) \right].$

• $e^{x^2}(1 + 2x^2)$ is constant with respect to $y$,
• The derivative of $y$ with respect to $y$ is $1$.

Thus:

$f_{xy} = e^{x^2} (1 + 2x^2).$

Simplify:

$f_{xy} = e^{x^2} (2x^2 + 1).$

Factor out $y$ where necessary (if considering $f_{xy}$ in this form).

Step 4: Compare with the given $f_{xy}$

The problem states:

$f_{xy} = y e^{x^2} (2x^2 + 1).$

Here, there seems to be an extra $y$ in the provided $f_{xy}$, which is incorrect. The correct mixed derivative is:

$f_{xy} = e^{x^2} (2x^2 + 1).$

Final Answer

• The given $f_x$ is True.
• The given $f_{xy}$ is False because of the extra $y$.

Clarification or Follow-Up?

Do you need further clarification or details on the steps? Let me know!

Related Questions:

1. How do you compute higher-order derivatives for $f(x,y) = xye^{x^2}$?
2. What is the mixed derivative theorem (Schwarz's theorem) for partial derivatives?
3. How does the product rule extend to multiple variables?
4. How can you verify partial derivatives using explicit differentiation?
5. What is the importance of mixed partial derivatives in physics or engineering?

Tip: Always check mixed partial derivatives carefully for symmetry, as they can reveal important properties of the function.