If f and g are differentiable in a show that $$ is also differentiable in a Suppose that $f,g: U\to \mathbb{K}$ where $U$ is an open set around $a$ are differentiable in $a$. Show that $<f,g>: U \to \mathbb{K}$ where $<f,g>(x) = <f(x),g(x)>$ is also differentiable i $a$.
I know $f$ is differentiable in $a$ iff there is a $T \in L(\mathbb{K},E)$ where $E$ is a vector space such that
\begin{align}
\lim_{x \to a} \frac{f(x)-f(a)-T(x-a)}{\Vert x-a \Vert} = 0
\end{align}
Here I claim that $<f,g>=<f',g>+<f,g'>$.
I think I have to show that $<f',g>+<f,g'>$ is linear, right? But I don't know how to show that either.Because for $\lambda\in \mathbb{K}$
$$<f',g>+<f,g'>(\lambda x)=<f'(\lambda x),g(\lambda x)> + <f(\lambda x),g'(\lambda x)>=\lambda<f'(x),g(\lambda x)>+\lambda<f(\lambda x),g'(x)>$$
But I don't know how to do the rest. Would someone please help? I have an exam tomorrow and I'm in a kind of hurry.
 A: By assumption, you can write that
\begin{align}
f(x) &= f(a) + T_f(x-a) + \varepsilon_f (x; a) \\
g(x) &= g(a) + T_g(x-a) + \varepsilon_g (x; a)
\end{align}
where $\varepsilon_f, \varepsilon_g$ vanish at $a$ (i.e. as $x \to a$) faster than $\| x - a \|$. Now, write
\begin{align}
\langle f(x), g(x) \rangle &= \langle f(a) + T_f(x-a) + \varepsilon_f (x; a), g(a) + T_g(x-a) + \varepsilon_g (x; a) \rangle \\
&= \langle f(a), g(a) \rangle + \langle T_f(x-a), g(a) \rangle + \langle f(a), T_g(x-a) \rangle + E_{f,g} (x; a)
\end{align}
where $E_{f,g} (x; a)$ can be written out explicitly, and vanishes at $a$ faster than $\| x - a \|$ (this is worth checking).
So, for $h = \langle f, g \rangle$, you can write that
\begin{align}
h(x) &= h(a) +  \langle T_f(x-a), g(a) \rangle + \langle f(a), T_g(x-a) \rangle + E_{f,g} (x; a) \\
&= h(a) + \langle T_f^*g(a) + T_g^*f(a), x-a \rangle + E_{f,g} (x; a).
\end{align}
It is thus natural to write $T_h = \left( T_f^*g(a) + T_g^*f(a) \right)^*, \varepsilon_h(x;a) = E_{f,g} (x; a)$ to obtain a similar expression to before:
\begin{align}
h(x) &= h(a) + T_h(x-a) + \varepsilon_h (x; a)
\end{align}
where $\varepsilon_h$ vanishes at $a$ faster than $\| x - a \|$, which is precisely what is needed to prove the differentiability of $h$ at $a$.
