Question

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

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

Hint:

For \(u=e^{xy}\), remember that \(\frac{\partial}{\partial x}(e^{xy})=y e^{xy}\) and \(\frac{\partial}{\partial y}(e^{xy})=x e^{xy}\).

Answer 1

The Correct Option is B

We are given: \[ u=e^{xy}. \] First find \[ \frac{\partial^2u}{\partial x^2}. \] Differentiate \(u\) with respect to \(x\): \[ \frac{\partial u}{\partial x}=y e^{xy}. \] Now differentiate again with respect to \(x\): \[ \frac{\partial^2u}{\partial x^2}=y\cdot y e^{xy}. \] \[ \frac{\partial^2u}{\partial x^2}=y^2e^{xy}. \] Now find \[ \frac{\partial^2u}{\partial y^2}. \] Differentiate \(u\) with respect to \(y\): \[ \frac{\partial u}{\partial y}=x e^{xy}. \] Differentiate again with respect to \(y\): \[ \frac{\partial^2u}{\partial y^2}=x\cdot x e^{xy}. \] \[ \frac{\partial^2u}{\partial y^2}=x^2e^{xy}. \] Therefore, \[ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} = y^2e^{xy}+x^2e^{xy}. \] \[ =(x^2+y^2)e^{xy}. \] Now substitute \((x,y)=(1,1)\): \[ =(1^2+1^2)e^{1\cdot 1}. \] \[ =(1+1)e. \] \[ =2e. \] Hence, the required value is \[ 2e. \]