Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function. If \( g(u, v) = f(u^2 - v^2) \), then
\[ \frac{\partial^2 g}{\partial u^2} + \frac{\partial^2 g}{\partial v^2} = \]

• A. \( 4(u^2 - v^2) f''(u^2 - v^2) \)
• B. \( 2f'(u^2 - v^2) + 4(u^2 - v^2) f''(u^2 - v^2) \)
• C. \( 4(u^2 + v^2) f''(u^2 - v^2) \)
• D. \( 2(u - v) f''(u^2 - v^2) \)

Hint:

For second derivatives of composite functions, use the chain rule and simplify carefully.

Answer 1

The Correct Option is C

Step 1: Finding the first derivatives.
To compute \( \frac{\partial^2 g}{\partial u^2} \) and \( \frac{\partial^2 g}{\partial v^2} \), we first find the first derivatives. From the definition of \( g(u, v) \): \[ g(u, v) = f(u^2 - v^2), \] we calculate the first derivative of \( g(u, v) \) with respect to \( u \): \[ \frac{\partial g}{\partial u} = 2u f'(u^2 - v^2). \] Similarly, the first derivative of \( g(u, v) \) with respect to \( v \) is: \[ \frac{\partial g}{\partial v} = -2v f'(u^2 - v^2). \]
Step 2: Finding the second derivatives.
Now, we take the second derivatives: \[ \frac{\partial^2 g}{\partial u^2} = 2f'(u^2 - v^2) + 4u^2 f''(u^2 - v^2), \] \[ \frac{\partial^2 g}{\partial v^2} = 2f'(u^2 - v^2) - 4v^2 f''(u^2 - v^2). \]
Step 3: Adding the second derivatives.
Adding the second derivatives: \[ \frac{\partial^2 g}{\partial u^2} + \frac{\partial^2 g}{\partial v^2} = 4(u^2 - v^2) f''(u^2 - v^2) + 2f'(u^2 - v^2). \]
Step 4: Conclusion.
Thus, the correct answer is (C).