Question

Let \( g: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function. Define \( f: \mathbb{R}^3 \to \mathbb{R} \) by \[ f(x, y, z) = g(x^2 + y^2 - 2z^2). \] Then \[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \] is equal to

• A. \( (4x^2 + 4y^2 - 42z^2) g''(x^2 + y^2 - 2z^2) \)
• B. \( (4x^2 + y^2 + 42z^2) g''(x^2 + y^2 - 2z^2) \)
• C. \( (4x^2 + y^2 - 22z^2) g''(x^2 + y^2 - 2z^2) \)
• D. \( (4x^2 + y^2 + 42z^2) g''(x^2 + y^2 - 2z^2) + 8g'(x^2 + y^2 - 2z^2) \)

Hint:

When dealing with composite functions, apply the chain rule to find partial derivatives, and remember to account for the second-order derivatives when summing.

Answer 1

The Correct Option is B

Step 1: Find the partial derivatives of \( f(x, y, z) \).
The function \( f(x, y, z) = g(x^2 + y^2 - 2z^2) \) is a composition of the function \( g \) and the expression \( x^2 + y^2 - 2z^2 \). The partial derivatives can be calculated using the chain rule.

Step 2: Apply the second-order derivatives.
Taking second-order derivatives with respect to \( x, y, \) and \( z \), we get: \[ \frac{\partial^2 f}{\partial x^2} = 4g''(x^2 + y^2 - 2z^2), \quad \frac{\partial^2 f}{\partial y^2} = 4g''(x^2 + y^2 - 2z^2), \quad \frac{\partial^2 f}{\partial z^2} = -8g''(x^2 + y^2 - 2z^2) + 8g'(x^2 + y^2 - 2z^2). \]
Step 3: Summing the terms.
Summing these gives the result: \[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = (4x^2 + y^2 + 42z^2) g''(x^2 + y^2 - 2z^2) + 8g'(x^2 + y^2 - 2z^2). \]
Step 4: Conclusion.
The correct answer is (D).