Assume $f,g: U \rightarrow \mathbb{R}^{m}$ are differentiable with total differential $df_{x}$ and $dg_{x}$, respectively and $U$ is an open set in $\mathbb{R}^{n}$.
Then the map $\lambda f+g$ is differentiable with total differential $d(\lambda f+g)_{x}=\lambda df_{x}+dg_{x}$ where $\lambda \in \mathbb{R}$.
Proof idea:
To show $\lambda f+g$ is differentiable I have to show that there exists a linear map $d(\lambda f+g)_{x}:\mathbb{R}^{n} \to \mathbb{R}^{m}$ such that
$$\lim \limits_{h \to 0} \frac{\lambda f(x+h)+g(x+h)-\lambda f(x)-g(x)-d(\lambda f+g)_{x}}{\lVert h \rVert}=0$$ (i.e. the error term decays sublinearly).
From linear algebra I know that if $S(v)$ and $T(v)$ are linear maps from one vector space $V$ to another vector space $W$ then their sum is defined as $(S+T)(v)=S(v)+T(v)$. It is easy to show that $(S+T)(v)$ is again a linear map from $V$ to $W$. Similarly, the product of a linear map with a scalar, defined as $(cS)(v)=cS(v)$, can be shown to be a linear map.
So consider $d(\lambda f+g)_{x}=\lambda df_{x}+dg_{x}$. Using the properties about linear maps given above this is clearly a linear map.
Plugging in yields
\begin{align} & \lim \limits_{h \to 0} \frac{\lambda f(x+h)+g(x+h)-\lambda f(x)-g(x)-d(\lambda f+g)_{x}}{\lVert h \rVert} \\ & =\lim \limits_{h \to 0} \frac{\lambda (f(x+h)-f(x)-df_{x})+g(x+h)-g(x)-dg_{x}}{\lVert h \rVert} \\ & =\lambda \lim \limits_{h \to 0} \frac{(f(x+h)-f(x)-df_{x})}{\lVert h \rVert}+\lim \limits_{h \to 0}\frac{g(x+h)-g(x)-dg_{x}}{\lVert h \rVert} \\ & =\lambda \times 0 + 0 = 0 \\ \end{align} where the last line follows from the differentiability of $f$ and $g$.
Can you please check if there are any issues with this proof? This might be an easy proof but I'm a bit worried about just assuming $d(\lambda f+g)_{x}=\lambda df_{x}+dg_{x}$ so any help is much appreciated.
Thanks very much!