TITLE: Derivative of sum of two functions is the sum of their derivatives.
QUESTION [1 upvotes]: Suppose $x_0 \in U \subseteq \mathbb{R}^d$, $U$ open, and $f,g : U \to \mathbb{R}^m$ differentiable at $x_0$, then $$D_{f + g} (x_0) = D_f(x_0) + D_g(x_0).$$ 

MY ATTEMPT
Put $r(x) = f(x) + g(x) $. We want to show that $D_r(x_0) = D_f (x_0) + D_g (x_0) $. We must show that 
$$ \lim_{h \to 0} \frac{ r(x_0 + h) - r(x_0) - D_f (x_0)(h) - D_g (x_0)(h)}{||h||} = 0.$$
But, notice
$$ \lim_{h \to 0} \frac{ r(x_0 + h) - r(x_0) - D_f (x_0)(h) - D_g (x_0)(h)}{||h||} = \lim_{h \to 0} \frac{ f(x_0 + h) + g(x_0+h) - f(x_0)-g(x_0) -  D_f (x_0)(h) - D_g (x_0)(h)}{||h||} $$
But, the limit above is $0$ by hypothesis. Then, we must have that $D_r(x_0) = D_f (x_0) + D_g (x_0)$.
Is this correct?

REPLY [3 votes]: Yes! Sure! In fact if $f(x)$ and $g(x)$ are two differentiable functions in $[a,b]$, and $r(x)=f(x)+g(x)$, then in $[a,b]$, $D[r(x)] = D[f(x)]+D[g(x)]$. In fact 
$$D[f(x_0)] = \lim_{\epsilon \to 0} \frac{f(x_0+ \epsilon)-f(x_0)}{\epsilon}$$
and 
$$D[g(x_0)] = \lim_{\epsilon \to 0} \frac{g(x_0+ \epsilon)-g(x_0)}{\epsilon}$$ 
so
$$D[f(x_0)] + D[g(x_0)] = \lim_{\epsilon \to 0} \frac{f(x_0+ \epsilon)-f(x_0)}{\epsilon} + \lim_{\epsilon \to 0} \frac{g(x_0+ \epsilon)-g(x_0)}{\epsilon} = \lim_{\epsilon \to 0} \frac{f(x_0+ \epsilon)-f(x_0)+g(x_0+ \epsilon)-g(x_0)}{\epsilon} = \lim_{\epsilon \to 0} \frac{r(x_0+ \epsilon)-r(x_0)}{\epsilon}=D[r(x_0)] \\\implies D_{f + g} (x_0) = D_f(x_0) + D_g(x_0)$$
{"set_name": "stack_exchange", "score": 1, "question_id": 700152}