How many total second-order partial derivatives are there for the
function f(x, y, z) = 3x2yz4 + 2xz cos(xy)? How many of them are distinct?
Calculate ∂ 4 f . ∂z2∂x∂y

To determine the total number of second-order partial derivatives for a function of three variables \( f(x, y, z) \), we use the fact that we can take derivatives with respect to each variable twice. In general, for a function of \( n \) variables, the number of second-order partial derivatives is given by the combination formula for choosing 2 items from \( n \) (with repetition), which is:

\[ \binom{n+2-1}{2} = \binom{n+1}{2} = \frac{(n+1)n}{2} \]

For \( n = 3 \) (since we have three variables: \( x, y, z \)):

\[ \binom{3+1}{2} = \binom{4}{2} = \frac{4 \cdot 3}{2} = 6 \]

Thus, there are 6 second-order partial derivatives. These are:

1. \( \frac{\partial^2 f}{\partial x^2} \)
2. \( \frac{\partial^2 f}{\partial y^2} \)
3. \( \frac{\partial^2 f}{\partial z^2} \)
4. \( \frac{\partial^2 f}{\partial x \partial y} \) (or \( \frac{\partial^2 f}{\partial y \partial x} \))
5. \( \frac{\partial^2 f}{\partial x \partial z} \) (or \( \frac{\partial^2 f}{\partial z \partial x} \))
6. \( \frac{\partial^2 f}{\partial y \partial z} \) (or \( \frac{\partial^2 f}{\partial z \partial y} \))

Since mixed partial derivatives are equal if the function and its partials are continuous (Schwarz's theorem or Clairaut's theorem), the number of distinct second-order partial derivatives is 6.

Next, we calculate the mixed partial derivative \( \frac{\partial^4 f}{\partial z^2 \partial x \partial y} \).

Given \( f(x, y, z) = 3x^2yz^4 + 2xz \cos(xy) \):

1. **First partial derivative with respect to \( y \):**

\[ f_y = \frac{\partial}{\partial y} (3x^2yz^4 + 2xz \cos(xy)) = 3x^2z^4 + 2xz \cdot (-x \sin(xy)) = 3x^2z^4 - 2x^2z \sin(xy) \]

2. **Second partial derivative with respect to \( x \):**

\[ \frac{\partial}{\partial x} \left(3x^2z^4 - 2x^2z \sin(xy)\right) = 6xz^4 - 2z \sin(xy) \cdot 2x - 2x^2z \cos(xy) \cdot y \]
\[ = 6xz^4 - 4xz \sin(xy) - 2x^2z y \cos(xy) \]

3. **Third partial derivative with respect to \( z \):**

\[ \frac{\partial}{\partial z} \left(6xz^4 - 4xz \sin(xy) - 2x^2z y \cos(xy)\right) = 6x \cdot 4z^3 - 4x \sin(xy) - 2x^2 y \cos(xy) \]
\[ = 24xz^3 - 4x \sin(xy) - 2x^2 y \cos(xy) \]

4. **Fourth partial derivative with respect to \( z \):**

\[ \frac{\partial}{\partial z} \left(24xz^3 - 4x \sin(xy) - 2x^2 y \cos(xy)\right) = 24x \cdot 3z^2 - 0 - 0 \]
\[ = 72xz^2 \]

Thus, the fourth partial derivative \( \frac{\partial^4 f}{\partial z^2 \partial x \partial y} \) is:

\[ \frac{\partial^4 f}{\partial z^2 \partial x \partial y} = 72xz^2 \]

Would you like more details or have any questions about this problem? Here are five questions you might want to ask next:

1. How do you find the first-order partial derivatives of \( f(x, y, z) \)?
2. What is Schwarz's theorem and how is it applied?
3. Can you explain the combination formula used to determine the number of second-order partial derivatives?
4. How do higher-order partial derivatives differ in their calculation?
5. Can you demonstrate a different example involving mixed partial derivatives?

**Tip:** When dealing with partial derivatives, ensure to follow the order of differentiation correctly and remember the rules of differentiation apply similarly to multivariable functions as they do in single-variable calculus.

Learn how to calculate second-order partial derivatives and mixed partial derivatives for a multivariable function. Explore the combination formula used to determine the number of second-order partial derivatives and apply Schwarz's theorem. Detailed example included.

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