Question:easy

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

Show Hint

Differentiate with respect to \(x\) twice while treating \(y\) as a constant, then substitute the given point.
Updated On: Aug 7, 2026
  • \(0\)
  • \(1\)
  • \(2\)
  • \(3\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Split the function into two separate terms.
$f(x,y) = x^2y + 3y^2x$ has two terms: $x^2y$ and $3y^2x$. Differentiate each term separately with respect to $x$, since the derivative of a sum is the sum of the derivatives.

Step 2: Differentiate the first term twice with respect to $x$.
Term 1 is $x^2y$. Treating $y$ as a constant coefficient:
First derivative: $\frac{\partial}{\partial x}(x^2y) = 2xy$.
Second derivative: $\frac{\partial}{\partial x}(2xy) = 2y$.

Step 3: Differentiate the second term twice with respect to $x$.
Term 2 is $3y^2x$, where $x$ appears only to the first power with coefficient $3y^2$:
First derivative: $\frac{\partial}{\partial x}(3y^2x) = 3y^2$.
Second derivative: $\frac{\partial}{\partial x}(3y^2) = 0$, since $3y^2$ has no $x$ left in it.

Step 4: Add the two second derivatives together.
\[ \frac{\partial^2 f}{\partial x^2} = 2y + 0 = 2y \]

Step 5: Substitute $x=1, y=1$ and pick the matching option.
\[ \left.\frac{\partial^2 f}{\partial x^2}\right|_{(1,1)} = 2(1) = 2 \]
This matches option (C).

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