Showing $h(x)=\langle f(x),g(x)\rangle$ where $f,g:\Bbb R^n\to\Bbb R^m$ are differentiable, is also differentiable 
I would like to prove $h:\Bbb R^n\to\Bbb R, h(x)=\langle f(x),g(x)\rangle,$ where $f,g:\Bbb R^n\to\Bbb R^m$ are differentiable functions, is also differentiable.

I've read this post, where the exercise is to compute $\frac{\partial h}{\partial v}(c)$ for an arbitrary $v\in\Bbb R^n.$ There, it is shown: $$\frac{\partial h}{\partial v}(c)=\left\langle\frac{\partial f}{\partial v}(c),g(c)\right\rangle+\left\langle f(c),\frac{\partial g}{\partial v}(c)\right\rangle.$$
I wanted to use the result and test the linear functional with the matrix
$\begin{aligned}L&=\begin{bmatrix} \left\langle\frac{\partial f}{\partial x_1}(c),g(c)\right\rangle+\left\langle f(c),\frac{\partial g}{\partial x_1}(c)\right\rangle&\ldots&\left\langle\frac{\partial f}{\partial x_n}(c),g(c)\right\rangle+\left\langle f(c),\frac{\partial g}{\partial x_n}(c)\right\rangle\end{bmatrix}\\&=\begin{bmatrix}\left\langle g(c),\frac{\partial f}{\partial x_1}(c)\right\rangle&\ldots&\left\langle g(c),\frac{\partial f}{\partial x_n}(c)\right\rangle\end{bmatrix}+\begin{bmatrix}\left\langle f(c),\frac{\partial g}{x_1}(c)\right\rangle&\ldots&\left\langle 
f(c),\frac{\partial g}{\partial x_n}(c)\right\rangle\end{bmatrix}\\&= g(c)^T Df(c)+f(c)^TDg(c).\end{aligned}$
as a candidate for the differential.
In order for $h$ to be differentiable, we need: $$\lim_{x\to c}\frac{\|\langle f(x),g(x)\rangle-\langle f(c),g(c)\rangle-(g(c)^TDf(c)+f(c)^TDg(c))(x-c)\|}{\|x-c\|}=0$$
We might write $(g(c)^TDf(c)+f(c)^TDg(c))(x-c)=\langle g(c)^TDf(c)+f(c)^TDg(c),x-c\rangle,$
but I got stuck. How should I proceed?
 A: Let $h:=x-c$. Let also $\varepsilon_1(h):=f(x)-f(c)-Df(c)h$, and $\varepsilon_2(h):=g(x)-g(c)-Dg(c)h$. Because $f$ and $g$ are differentiable at $c$, we have $\lim_{\|h\|\to0}\frac{\varepsilon_i(h)}{\|h\|}=0$ for $i=1,2$. Now, by bilinearity,
\begin{align*}
f(x)^Tg(x)
&=\bigl(f(c)+ Df(c)h+\varepsilon_1(h)\bigr)^T\bigl(g(c)+ Dg(c)h+\varepsilon_2(h)\bigr)\\
&=f(c)^Tg(c)+\underbrace{\left(f(c)^TDg(c)+g(c)^TDf(c)\right)h}_{\ell(c)(h)}
+\varepsilon_3(h),
\end{align*}
where $\varepsilon_3(h):=\varepsilon_1(h)^T\bigl(g(c)+Dg(c)h\bigr)+\varepsilon_2(h)^T\bigl(f(c)+Df(c)h\bigr)+\varepsilon_1(h)^T\varepsilon_2(h)$ is also such that
$\lim_{\|h\|\to0}\frac{\varepsilon_3(h)}{\|h\|}=0$ by Cauchy-Schwarz inequality. Moreover, $\ell(c)$ is clearly a linear map, and it is also bounded (again, by Cauchy-Schwarz inequality).
This shows that $x\mapsto f(x)^Tg(x)$ is differentiable at $c$ with differential $\ell(c)$.
A: Don't confuse the differential with the gradient.
Using Leibniz we get for $h(p)=\langle f(p),g(p)\rangle$ the differential
$$d_ph(v)=\langle d_pf(v),g(p)\rangle+
\langle f(p),d_pg(v)\rangle.$$
In terms of the Jacobian $J_f$ of $f$ and $J_g$ of $g$ we get
$$
\begin{align}
\langle J_f(p)v,g(p)\rangle+
\langle f(p),J_g(p)(v)\rangle&=
\langle v,J^T_f(p)g(p)+\langle J^T_g(p)f(p),v\rangle\\
&=\langle J^T_f(p)g(p)+J^T_g(p)f(p),v \rangle,
\end{align}
$$
hence
$$\nabla \langle f(p),g(p)\rangle
=J^T_f(p)g(p)+J^T_g(p)f(p)
$$
