We are given the function:

$f(x, y) = \sin(2x^2y^2 + 6x^3y) + 10$

We are asked to show that the mixed partial derivatives $f_{xy}(x, y)$ and $f_{yx}(x, y)$ are equal, i.e., that the mixed second derivatives commute.

Steps:

1. Find the first partial derivative of $f$ with respect to $x$: $f_x(x, y) = \frac{\partial}{\partial x} \left( \sin(2x^2y^2 + 6x^3y) + 10 \right)$ Using the chain rule, we differentiate the sine function, and the constant 10 disappears (since the derivative of a constant is zero). For the sine part, the derivative is the cosine of the argument multiplied by the derivative of the inner function:

   $f_x(x, y) = \cos(2x^2y^2 + 6x^3y) \cdot \frac{\partial}{\partial x} \left( 2x^2y^2 + 6x^3y \right)$

   The derivative of the inner function with respect to $x$ is:

   $\frac{\partial}{\partial x}(2x^2y^2) = 4xy^2$ $\frac{\partial}{\partial x}(6x^3y) = 18x^2y$ So: $f_x(x, y) = \cos(2x^2y^2 + 6x^3y) \cdot (4xy^2 + 18x^2y)$

2. Find the second partial derivative with respect to $y$, i.e., $f_{xy}(x, y)$: Now differentiate $f_x(x, y)$ with respect to $y$: $f_{xy}(x, y) = \frac{\partial}{\partial y} \left( \cos(2x^2y^2 + 6x^3y) \cdot (4xy^2 + 18x^2y) \right)$ This requires using the product rule, as we have two functions of $y$.

   Let me compute this expression step-by-step.The mixed partial derivatives $f_{xy}(x, y)$ and $f_{yx}(x, y)$ are:

$f_{xy}(x, y) = (18x^2 + 8xy) \cos(6x^3y + 2x^2y^2) - (6x^3 + 4x^2y)(18x^2y + 4xy^2) \sin(6x^3y + 2x^2y^2)$

$f_{yx}(x, y) = (18x^2 + 8xy) \cos(6x^3y + 2x^2y^2) - (6x^3 + 4x^2y)(18x^2y + 4xy^2) \sin(6x^3y + 2x^2y^2)$

Since $f_{xy}(x, y) = f_{yx}(x, y)$, this confirms that the mixed second partial derivatives are equal, as required.

Would you like to explore any details further or have any questions about this?

Relative questions:

1. How is the product rule applied to partial derivatives in multivariable calculus?
2. Can you explain how the chain rule was used in this problem?
3. What conditions guarantee that mixed partial derivatives are equal?
4. What is the significance of Clairaut's theorem in this context?
5. How would this result change if a different function were given?

Tip:

When calculating mixed partial derivatives, always check if the function is smooth and continuous across its domain to apply Clairaut's theorem effectively.