I guess F is not a conservative vector field.
$$F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}$$
$$\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=cosy$$
$$\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x$$
$$cosy\neq e^x$$

Question

I guess F is not a conservative vector field. 
$$F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}$$
$$\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=cosy$$
$$\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x$$
$$cosy\neq e^x$$

anonymous · Answer

The second one, I am not sure what you defined as which, but just to make sure. You are taking the partial derivative of e^x cosy ? with respect to x?

anonymous · Answer

$$\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=e^x cosy$$

anonymous · Answer

treat the respective values as constants for partial derivatives.

anonymous · Answer

exactly @91

anonymous · Answer

$$\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x cos y$$

so there are conservative

anonymous · Answer

it is a conservative vector field.

anonymous · Answer

constant multplied to variable don't go away

anonymous · Answer

Oh ok. Thanks 
in the previous example that I did I got 
$$\frac{\partial}{\partial{y}}|2x-3y|=-3$$
Which lead me to believe that the constant went away

anonymous · Answer

only if they are by themselve

anonymous · Answer

Makes sense. Thank you both!