Question

Let \(f\) be a function of real variables \(x\) and \(y\), defined as:
\[ f(x,y) = x^2y + 3y^2x \]
The value of \( \frac{\partial^2 f}{\partial x^2} \) at \(x=1, y=1\) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

Differentiate with respect to \(x\) twice while treating \(y\) as a constant, then substitute the given point.

Answer 1

The Correct Option is C

Step 1: Find the first partial derivative with respect to \(x\).
Treat \(y\) as a constant and differentiate \(f(x,y) = x^2y + 3y^2x\) with respect to \(x\).
\[ \frac{\partial f}{\partial x} = 2xy + 3y^2 \]

Step 2: Differentiate again with respect to \(x\) to get the second partial derivative.
Differentiate \(\frac{\partial f}{\partial x} = 2xy + 3y^2\) with respect to \(x\) again, still treating \(y\) as constant. The term \(3y^2\) has no \(x\) left in it, so its derivative with respect to \(x\) is \(0\).
\[ \frac{\partial^2 f}{\partial x^2} = 2y \]

Step 3: Substitute the given point.
At \(x=1, y=1\):
\[ \frac{\partial^2 f}{\partial x^2} = 2(1) = 2 \]

Step 4: Check why the other options are wrong.
Option (A) 0 would only occur if the \(x^2y\) term were mistakenly dropped before differentiating twice. Option (B) 1 does not match any natural intermediate result in this calculation. Option (D) 3 would come from confusing \(\frac{\partial^2 f}{\partial x^2}\) with \(\frac{\partial^2 f}{\partial y^2}\); differentiating \(3y^2x\) twice with respect to \(y\) gives \(6x\), which is \(6\) at \(x=1\), not \(3\), so even that mix-up does not give 3. The careful calculation above gives \(2\).

Final Answer:
\[ \boxed{2} \]