Question

If \(u=e^{xy}\), then the value of \[ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} \] at \((1,1)\) is

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

Hint:

For \(e^{xy}\), remember chain rule carefully in partial differentiation.

Answer 1

The Correct Option is B

Step 1: Given: \[ u=e^{xy} \]

Step 2: \[ \frac{\partial u}{\partial x}=ye^{xy} \] \[ \frac{\partial^2u}{\partial x^2}=y^2e^{xy} \]

Step 3: \[ \frac{\partial u}{\partial y}=xe^{xy} \] \[ \frac{\partial^2u}{\partial y^2}=x^2e^{xy} \]

Step 4: Add: \[ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} = (y^2+x^2)e^{xy} \]

Step 5: At \((1,1)\): \[ (1^2+1^2)e^{1\cdot 1}=2e \] \[ \boxed{2e} \]