Show that $\langle f,g \rangle'(x_0)(h)=\langle g(x_0),f'(x_0)(h)\rangle + \langle f(x_0), g'(x_0)(h) \rangle$ If $f,g:A\rightarrow\mathbb{R}^m$ are both differentiable in $x_0\in A\subseteq \mathbb{C}^n$, then $\langle f,g \rangle: A\rightarrow \mathbb{R}$ is differentiable in $x_0$ with $\langle f,g \rangle'(x_0)(h)=\langle g(x_0),f'(x_0)(h)\rangle + \langle f(x_0), g'(x_0)(h) \rangle$.

I have the following proven statement available: 
$m:\mathbb{R}^m\times\mathbb{R}^m\rightarrow\mathbb{R},(x,y)\mapsto\langle x,y\rangle$ is differentiable in all $(x_0,y_0)\in\mathbb{R}^m\times\mathbb{R}^m$ with $m'(x_0,y_0)(u,v)=\langle y_0,u \rangle + \langle x_0,v \rangle$.

I tried for a few hours at this, but I really can't handle this. I know it's just manipulating definitions and I even know that I need to simply apply the chain rule, but I fail at writing it down. Can anybody point me to a proof of this, if possible or illustrate the main steps?
 A: I'll tell you something even better: if $B\colon \Bbb R^n \times \Bbb R^m \to \Bbb R^k$ is bilinear, then $B$ is differentiable and the total derivative is given by $$DB(x,y)(h,k) = B(x,k) + B(h,y).$$If you forget the very frequent instances where $B$ is symmetric and $n=m$, you see that the above is the only possible formula. The proof uses that you can get a constant $C$ such that $\|B(x,y)\| \leq C\|x\|\|y\|$ for all $x \in \Bbb R^n$ and $y \in \Bbb R^m$, and you use that to check the formal definition. I think you can do this on your own, assuming the existence of that $C$, at least (to find $C$, use bases). 
Ok, that said, we consider $x \mapsto \langle f(x),g(x)\rangle$. By the chain rule we get $$\begin{align} D(\langle\cdot,\cdot\rangle\circ (f,g))(x_0)(h) &= D(\langle\cdot,\cdot\rangle)(f(x_0),g(x_0)) \circ (Df(x_0),Dg(x_0))(h) \\ &= D(\langle\cdot,\cdot\rangle)(f(x_0),g(x_0))(Df(x_0)(h), Dg(x_0)(h)) \\ &= \langle f(x_0), Dg(x_0)(h)\rangle+ \langle Df(x_0)(h), g(x_0)\rangle,\end{align}$$as wanted.
