Show that $$\frac{\partial(fg)}{\partial z}=\frac{\partial(f)}{\partial z}g+f\frac{\partial(g)}{\partial z}$$ where $f,g$ are $\mathbb{R}$-differentiable.
Proof: Suppose that $f,g$ are $\mathbb{R}$-differentiable. From Calculus, we know that $fg$ is also $\mathbb{R}$-differentiable. i.e. For $x\in\mathbb{R}$, $$\dfrac{\partial(fg)}{\partial x}=\dfrac{\partial(f)}{\partial x}g+f\dfrac{\partial(g)}{\partial x}$$
Now my gut says that proving this for the complex ($\mathbb{C}$) case isn't the same as the real ($\mathbb{R}$) proof. Am I right to trust my gut? How should I continue with this problem?
Edit: Suppose that $f,g$ are $\mathbb{R}$-differentiable. From Calculus, we know that $fg$ is also $\mathbb{R}$-differentiable. i.e. $f'$ and $g'$ exists for all $x\in\mathbb{R}$. Recall that $$\dfrac{\partial(fg)}{\partial z}=\dfrac{1}{2}\left(\dfrac{\partial(fg)}{\partial x}-i\dfrac{\partial(fg)}{\partial y}\right)=\dfrac{1}{2}\left(\dfrac{\partial(fg)}{\partial x}+\dfrac{1}{i}\dfrac{\partial(fg)}{\partial y}\right).$$ With $f=u+iv$, $g=\alpha+i\beta$ for real-valued $u,v,\alpha,\beta$, then $fg=\alpha u-\beta v+i(\beta u+\alpha v)$, so \begin{equation*} \begin{aligned} \dfrac{\partial(fg)}{\partial x} & = \dfrac{\partial}{\partial x}(u\alpha-v\beta) +i\dfrac{\partial}{\partial x}(u\beta+v\alpha) \\ & = \alpha \dfrac{\partial u}{\partial x} + u \dfrac{\partial \alpha}{\partial x} -\beta \dfrac{\partial v}{\partial x} - v \dfrac{\partial \beta}{\partial x} + i u \dfrac{\partial \beta}{\partial x} + i \beta \dfrac{\partial u}{\partial x} + i \alpha \dfrac{\partial v}{\partial x} + i v \dfrac{\partial \alpha}{\partial x}\\ & = \dfrac{\partial u}{\partial x}(\alpha+i\beta) + \dfrac{\partial \alpha}{\partial x}(u+iv) + \dfrac{\partial v}{\partial x}(-\beta+i\alpha) + \dfrac{\partial \beta}{\partial x}(-v+iu) \end{aligned} \end{equation*} Note $-\beta+i\alpha=i^2\beta+i\alpha=i(\alpha+i\beta)$ and similarly $-v+iu=i(u+iv)$. Then, \begin{equation*} \begin{aligned} \dfrac{\partial(fg)}{\partial x} & = \dfrac{\partial u}{\partial x} g + \dfrac{\partial \alpha}{\partial x} f + i\dfrac{\partial v}{\partial x} g + i\dfrac{\partial \beta}{\partial x} f \\ & = \left(\dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right) g + \left(\dfrac{\partial \alpha}{\partial x} + i\dfrac{\partial \beta}{\partial x} \right) f \\ & = g \dfrac{\partial}{\partial x} (u+iv) + f \dfrac{\partial}{\partial x} (\alpha + i\beta) \\ & = g \dfrac{\partial f}{\partial x} + f \dfrac{\partial g}{\partial x} \end{aligned} \end{equation*}
Similarly, we can find $$\frac{\partial(fg)}{\partial y} = g \frac{\partial f}{\partial y} + f \frac{\partial g}{\partial y} $$ Thus, \begin{equation*} \begin{aligned} \frac{\partial(fg)}{\partial z} & =\frac{1}{2}\left(\frac{\partial(fg)}{\partial x}+\frac{1}{i}\frac{\partial(fg)}{\partial y}\right) \\ & =\dfrac{1}{2} \left(g \dfrac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} +\dfrac{1}{i}\left(g \frac{\partial f}{\partial y} + f \frac{\partial g}{\partial y}\right)\right) \\ & =\dfrac{1}{2} \left(g\left(\frac{\partial f}{\partial x} +\frac{1}{i} \frac{\partial f}{\partial y}\right) + f\left(\frac{\partial g}{\partial x} +\frac{1}{i}\frac{\partial g}{\partial y}\right) \right) \\ \end{aligned} \end{equation*}
How do I continue from here?
With $f=u+iv$, $g=\alpha+i\beta$ for real-valued $u,v,\alpha,\beta$, then $fg=\alpha u-\beta v+i(\beta u+\alpha v)$, so \begin{align*} \dfrac{\partial(fg)}{\partial x}=\dfrac{\partial}{\partial x}(\alpha u-\beta v)+i\dfrac{\partial}{\partial x}(\beta u+\alpha v)=\cdots \end{align*} and in a routine way also, \begin{align*} g\dfrac{\partial f}{\partial x}=(\alpha+i\beta)\dfrac{\partial}{\partial x}(u+iv)=\cdots \end{align*} after a messy simplification, the result follows.
Note that \begin{align*} \dfrac{\partial}{\partial z}(fg)=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}(fg)+\dfrac{1}{i}\dfrac{\partial}{\partial y}(fg)\right). \end{align*}