Why is $f(x)g'(x)+f'(x)g(x)$ a linear map? Assume $f:E \rightarrow \mathbb{R}^m$ and $g:E \rightarrow \mathbb{R}$, and $E \in \mathbb{R}^n$ is open. Assume $x \in E$, and $f$ and $g$ are differentiable at $x$. To get the derivative of $f(x)g(x)$, I get a product rule version, namely $f'(x)g(x)+g'(x)f(x)$. 
According to the definition of a derivative (In Rudin's Principles of Mathematical Analysis), $f'(x)g(x)+g'(x)f(x)$ has to be a linear map. But why is it a linear map? Doesn't that imply we need g and f to be linear as well?

To sort this out more carefully, let's consider the derivative at a point $x$ in the direction $v\in\Bbb R^n$. Then $g(x)\big(f'(x)v\big)$ is the product of a scalar with a vector in $\Bbb R^m$. Similarly, $\big(g'(x)v\big)f(x)$ is the product of a scalar with a vector in $\Bbb R^m$. Both expressions are linear in $v$, so we end up with a linear map from $\Bbb R^n$ to $\Bbb R^m$.
