Question:hard

If f(x, y) \(=\) x\(^3\)y \(+\) xy\(^3\) \(+\) e\(^{xy^2}\), then f\(_y\)(1, 1) is

Show Hint

When taking partial derivatives:
- Keep the non-differentiating variable strictly constant in your mind.
- Verify the chain rule component: \( \frac{d}{du}(e^u) = e^u \frac{du}{dy} \).
Double check simple arithmetic to avoid losing easy marks.
Updated On: Jul 3, 2026
  • 2 \(+\) 2e
  • 2 \(+\) 4e
  • 4 \(+\) 2e
  • 4 \(+\) 4e
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Differentiate term by term with respect to y, plugging in numbers as soon as each piece is ready.
For \( f(x,y) = x^3y + xy^3 + e^{xy^2} \), differentiating \( x^3y \) with respect to \( y \) gives \( x^3 \), which at \( x = 1 \) is simply \( 1 \).

Step 2: Handle the second term the same way.
Differentiating \( xy^3 \) with respect to \( y \) gives \( 3xy^2 \), which at \( x=1, y=1 \) becomes \( 3(1)(1) = 3 \).

Step 3: Handle the exponential term using the chain rule.
Differentiating \( e^{xy^2} \) with respect to \( y \) gives \( e^{xy^2}\cdot 2xy \), and at \( x=1, y=1 \) the exponent \( xy^2 = 1 \), so this becomes \( e^1 \cdot 2(1)(1) = 2e \).

Step 4: Final Answer.
Adding the three evaluated pieces:
\[ f_y(1,1) = 1 + 3 + 2e = \boxed{4 + 2e} \]
Therefore, option (C) is correct.
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