Question

If \(z = 3x^2y - x \sin xy\), then:

• A. \(\frac{\partial^2 z}{\partial y \partial x} > \frac{\partial^2 z}{\partial x \partial y}\)
• B. \(\frac{\partial^2 z}{\partial y \partial x} < \frac{\partial^2 z}{\partial x \partial y}\)
• C. \(\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}\)
• D. \(\frac{\partial^2 z}{\partial y \partial x} = 0\)

Hint:

For any standard, well-behaved mathematical function composed of polynomials, exponentials, sines, and cosines, the mixed partial derivatives are always equal:
\[ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \]
You can select this option immediately without doing any differentiation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem relates to partial differentiation, specifically the equality of mixed second-order partial derivatives.
Clairaut's Theorem (or Schwarz's Theorem) governs when these mixed partial derivatives are equal.

Step 3: Detailed Explanation:
Let us state Clairaut's Theorem on mixed partial derivatives:
"If the function \(z = f(x, y)\) and its partial derivatives \(f_x\), \(f_y\), \(f_{xy}\), and \(f_{yx}\) are continuous in an open region, then at any point in that region:
\[ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y} \]
The given function is:
\[ z = 3x^2y - x \sin(xy) \]
This function is a combination of polynomials and trigonometric functions.
Since polynomial and trigonometric functions are infinitely differentiable and continuous everywhere, their derivatives of all orders are also continuous everywhere on \(\mathbb{R}^2\).
Therefore, Clairaut's Theorem holds true for this function.
Let us verify this by calculating both mixed partial derivatives explicitly:
First, find the first-order partial derivative with respect to \(x\), treating \(y\) as a constant:
\[ \frac{\partial z}{\partial x} = 6xy - \sin(xy) - x y \cos(xy) \]
Now, differentiate this with respect to \(y\), treating \(x\) as a constant:
\[ \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \left[ 6xy - \sin(xy) - xy\cos(xy) \right] \]
\[ \frac{\partial^2 z}{\partial y \partial x} = 6x - x\cos(xy) - x\cos(xy) + xy \cdot x\sin(xy) \]
\[ \frac{\partial^2 z}{\partial y \partial x} = 6x - 2x\cos(xy) + x^2 y\sin(xy) \quad \text{--- (Equation 1)} \]
Next, find the first-order partial derivative with respect to \(y\), treating \(x\) as a constant:
\[ \frac{\partial z}{\partial y} = 3x^2 - x^2 \cos(xy) \]
Now, differentiate this with respect to \(x\), treating \(y\) as a constant:
\[ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left[ 3x^2 - x^2 \cos(xy) \right] \]
Using the product rule:
\[ \frac{\partial^2 z}{\partial x \partial y} = 6x - 2x\cos(xy) - x^2 \left(-\sin(xy) \cdot y\right) \]
\[ \frac{\partial^2 z}{\partial x \partial y} = 6x - 2x\cos(xy) + x^2 y\sin(xy) \quad \text{--- (Equation 2)} \]
Comparing Equation 1 and Equation 2, we see that they are exactly equal.
This confirms the theorem and matches Option (C).

Step 4: Final Answer:
Therefore, the correct option is (C).