Form a function \(z=f(x,y)\) such that it satisfies \(f_x=\sin y + 1\) and \(f_y = x\cos y\)

\[ x \sin y+x \]

The anti-derivative of \(\cos\) is \(\sin\). Therfore, the term \(f_y\) term must integrate to \(x\sin y\) this matches what we see in \(f_x\) but without the \(x\) term. Because the term is not zero, the there must be an \(x\) infront of it before ther derivative WRT \(x\) was taken. We also need to have a term that goes to zero for \(f_y\) and one for \(f_x\). This is simply \(x\).

To verify:

\[ f(x,y) = x \sin y +x \\ f_x = \sin y + 1 \\ f_y=x\cos y \]

Thus, we have the correct formula. We do not need to use a constant c in this because we only need to find ONE function, not tall. More generally, we coudl write:

\[ f(x,y) = x \sin y +x + c \]