Final answer:
We verified Clairaut's Theorem for the function u=e^{xy}sin(y) by showing that the mixed partial derivatives with respect to x and y are indeed equal, which confirms the theorem's criteria for the given function.
Step-by-step explanation:
The student has asked to verify Clairaut's Theorem for a given function u=e^{xy}\sin(y). Clairaut's Theorem states that if a function u=f(x,y) has continuous second partial derivatives, then the mixed derivatives are equal, that is:
\[\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}\]
We first find the partial derivative of u with respect to x:
\[\frac{\partial u}{\partial x} = y e^{xy} \sin(y)\]
Then we take the partial derivative of this result with respect to y:
\[\frac{\partial^2 u}{\partial y \partial x} = e^{xy} \sin(y) + xy e^{xy} \cos(y)\]
Now, we reverse the order. We first find the partial derivative of u with respect to y:
\[\frac{\partial u}{\partial y} = xe^{xy} \sin(y) + e^{xy} \cos(y)\]
And then take the partial derivative of this result with respect to x:
\[\frac{\partial^2 u}{\partial x \partial y} = e^{xy} \sin(y) + xy e^{xy} \cos(y)\]
Both mixed derivatives are indeed equal, hence verifying Clairaut's Theorem for the given function u=e^{xy}\sin(y).