Question

Let \( g: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function. If \( f(x, y) = g(y) + x g'(y) \), then

• A. \( \dfrac{\partial f}{\partial x} + y \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial f}{\partial y} \)
• B. \( \dfrac{\partial f}{\partial y} + y \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial f}{\partial x} \)
• C. \( \dfrac{\partial f}{\partial x} + \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial f}{\partial y} \)
• D. \( \dfrac{\partial f}{\partial y} + x \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial f}{\partial x} \)

Hint:

When solving partial derivative problems, always compute derivatives step-by-step and match terms carefully; symmetry of mixed derivatives often simplifies verification.

Answer 1

The Correct Option is C

Step 1: Calculate the First Partial Derivatives

We first find the partial derivative of $f$ with respect to $x$ and $y$. Note that $g(y)$ and $g'(y)$ are treated as constants when differentiating with respect to $x$.

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( g(y) + xg'(y) \right)$$

$$\frac{\partial f}{\partial x} = 0 + 1 \cdot g'(y)$$

$$\frac{\partial f}{\partial x} = g'(y)$$

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( g(y) + xg'(y) \right)$$

$$\frac{\partial f}{\partial y} = g'(y) + x \cdot g''(y)$$

$$\frac{\partial f}{\partial y} = g'(y) + xg''(y)$$

Step 2: Calculate the Second Partial Derivative $\frac{\partial^2 f}{\partial x \partial y}$

Since $f$ is twice differentiable, the mixed partial derivatives $\frac{\partial^2 f}{\partial x \partial y}$ and $\frac{\partial^2 f}{\partial y \partial x}$ are equal (by Clairaut's theorem). We calculate the former:

$$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} \left( g'(y) + xg''(y) \right)$$

Note that $g'(y)$ and $g''(y)$ are treated as constants when differentiating with respect to $x$.

$$\frac{\partial^2 f}{\partial x \partial y} = 0 + 1 \cdot g''(y)$$

$$\frac{\partial^2 f}{\partial x \partial y} = g''(y)$$

The correct option is (C).