I have started differentiation very recently and by the looks of the proof of the Cauchy-Riemann equations in a polar form, I have difficulty converting the following:
$\frac{\partial v}{\partial y}$ $cos(Ф)$ - $\frac{\partial v}{\partial x}$ $cos(Ф)$ = $\frac{r}{1}$$\frac{\partial v}{\partial Ф}$
Well, I can try to factor out 1/r from the expression and get
$\frac{1}{r}$ ( $\frac{\partial v}{\partial y}$ $rcos(Ф)$ - $\frac{\partial v}{\partial x}$ $rcos(Ф)$)
but then I cannot equal it to ∂v/∂Ф. Then, I figured out if I could use the Chain Rule, but it takes me nowhere.
Can anybody explain this to me? I will be glad to any feedback.
Also, I am sorry my equations look little unprofessional; I am currently studying how to write these correctly.
Thank You!
$$ \begin{aligned} v(r, \Phi) &= v(x(r, \Phi), y(r, \Phi)) \Rightarrow \\ \frac{\partial v}{\partial \Phi} &= \frac{\partial v}{\partial x}\frac{\partial x}{\partial \Phi} + \frac{\partial v}{\partial y}\frac{\partial y}{\partial \Phi} = \\ &=\left[ \begin{array}{rl} x(r, \Phi) = r\cos(\Phi) \Rightarrow & \frac{\partial x}{\partial \Phi} = -r\sin(\Phi) \\ y(r, \Phi) = r\sin(\Phi) \Rightarrow & \frac{\partial y}{\partial \Phi} = r\cos(\Phi) \\ \end{array} \right] = \\ &= \frac{\partial v}{\partial x}\left(-r\sin(\Phi)\right) + \frac{\partial v}{\partial y}\left(r\cos(\Phi)\right) = \\ &=\frac{\partial v}{\partial y}r\cos(\Phi) - \frac{\partial v}{\partial x}r\sin(\Phi) \end{aligned} $$