If $f,g$ differentiable on $(a,b)$ can I say $f+g$ differentiable on $(a,b)$? I know that if $f$ and $g$ are differentiable at $a$, then $f+g$ is differentiable at $a$. 
But is it also true for the case of an open interval? For instance, if $f,g$ are differentiable on $(a,b)$, does $f+g$ also have to be differentiable on $(a,b)$?
 A: Yes this is true. That a function is differentiable on an open intervcal means exactly that it is differentiable at every points in the interval. So if $f$ and $g$ are differentiable at every point, then $f+g$ is also differentiable at every points.
A: Hint: For all $x\in (a,b),$ and for any real $h\ne0$ such that $x+h\in(a,b),$ we have that $$\begin{align}\frac{(f+g)(x+h)-(f+g)(x)}h &:=\frac{\left(f(x+h)+g(x+h)\right)-\left(f(x)+g(x)\right)}h\\ &=\frac{f(x+h)+g(x+h)-f(x)-g(x)}h\\ &=\frac{f(x+h)-f(x)}h+\frac{g(x+h)-g(x)}h.\end{align}$$ By assumption, $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}h$ and $\lim\limits_{h\to 0}\frac{g(x+h)-g(x)}h$ exist for all $x\in(a,b).$ So, what can we say about $\lim\limits_{h\to 0}\frac{(f+g)(x+h)-(f+g)(x)}h$ for all such $x$?
Note: The converse need not hold. Let $f$ be your favorite non-differentiable function and let $g=-f.$
A: Yes, let $f$ and $g$ be real functions in an open interval $(a,b)$ where $(a,b)$ is a subset of the real numbers. Then if $f$ and $g$ are differentiable at $a$, then $f + g$ is differentiable at $a$. 
Thus, $(f+g)'(a) = f'(a) + g'(a)$ is differentiable on $(a,b)$
In fact, this is also true for the following in an open interval.
1) $(\alpha f)'(a) = \alpha f'(a)$ where $\alpha$ is a real number
2) $(fg)'(a) = g(a)f'(a) + f(a)g'(a)$
3) $(\frac{f}{g})'(a) = \frac{g(a)f'(a) - f(a)g'(a)}{g^2} $
