How to differentiate Complex Fluid Potential I have $$F(z) = \phi + i\psi$$
Also, it is given that  $$u = \frac{\delta \phi}{\delta x}, \ \ v = \frac{\delta\phi}{\delta y}$$
and $$u = \frac{\delta \psi}{\delta x}, \ \ v = -\frac{\delta\psi}{\delta y}$$
Now I have to prove that $$\frac{dF}{dz} = u - iv$$
But I am getting terms like $\dfrac{dx}{dz}$ and $\dfrac{dy}{dz}$ in the differentiated term. How do I proceed?
EDIT:
As pointed out the function $\psi$ was wrongly defined by me. It's actually
$$u = \frac{\delta \psi}{\delta y}, \ \ v = -\frac{\delta\psi}{\delta x}$$
 A: Let's get some formulas straight in the first place. You have
$$F(z) = \phi + i\psi
$$
Also, it is given that
$$u = \frac{\partial \phi}{\partial x}, \ \ v = \frac{\partial \phi}{\partial y}
$$
And no, what you have instead is this:
$$u = \frac{\partial \psi}{\partial y}, \ \ v = -\frac{\partial \psi}{\partial x}
$$
Differentiate in the $x$-direction or in the $y$-direction - it doesn't matter (why?):
$$ \frac{dF}{dz} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial z}
   \qquad ; \qquad
   \frac{dF}{dz} = \frac{\partial F}{\partial y}\frac{\partial y}{\partial z}
$$
With 
$$ \frac{\partial z}{\partial x} = 1 \quad \Longrightarrow \quad \frac{\partial x}{\partial z} = 1/1 = 1 \qquad ; \qquad
 \frac{\partial z}{\partial y} = i \quad \Longrightarrow \quad \frac{\partial y}{\partial z} = 1/i = -i
$$
Hence in the $x$- or $y$-direction, take your favorite:
$$
\frac{dF}{dz} = 1 \left(\frac{\partial \phi}{\partial x} + i \frac{\partial \psi}{\partial x} \right) = 
-i \left( \frac{\partial \phi}{\partial y} + i \frac{\partial \psi}{\partial y}\right)
= u - i v
$$
More about this subject in Cauchy-Riemann Equations (PDF).
