Question

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

Always check if the function is continuous and has continuous partial derivatives.
Clairaut's theorem simplifies many problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For functions with continuous second-order partial derivatives,
Clairaut's theorem states that the mixed partial derivatives are equal.
That is, \(\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial^2 z}{\partial x \partial y}\).

Step 2: Key Formula or Approach:
Clairaut's theorem: If the second-order partial derivatives are continuous,
then the order of differentiation does not matter.

Step 3: Detailed Explanation:
Given \(z = 3x^2 y - x\sin(xy)\).
First, find \(\frac{\partial z}{\partial x}\): \[ \frac{\partial z}{\partial x} = 6xy - \sin(xy) - xy\cos(xy). \]
Now, find \(\frac{\partial}{\partial y}\) of the above: \[ \frac{\partial^2 z}{\partial y \partial x} = 6x - \left( x\cos(xy) \cdot x \right) - \left( x\cos(xy) + xy \cdot (-x\sin(xy)) \right). \]
Simplify: \[ = 6x - x^2\cos(xy) - x\cos(xy) + x^2 y \sin(xy). \]
Now, find \(\frac{\partial z}{\partial y}\): \[ \frac{\partial z}{\partial y} = 3x^2 - x^2\cos(xy). \]
Now, find \(\frac{\partial}{\partial x}\) of the above: \[ \frac{\partial^2 z}{\partial x \partial y} = 6x - \left( 2x\cos(xy) + x^2(-y\sin(xy)) \right). \]
Simplify: \[ = 6x - 2x\cos(xy) + x^2 y \sin(xy). \]
Comparing both results, they are equal.
Thus, the mixed partial derivatives are equal.
Hence, option (C) is correct.
This is a direct application of Clairaut's theorem.