I am stuck with proving the following statement. In fact, I am proving another assumption, and the proof of this would help me to proceed.
Assume that $f_1$ and $f_2$ are differentiable on the interval $(a,b)$. Then define $f=f_1+f_2$. Show that $f$ is differentiable on any point in that interval. So I guess I need to prove that the sum of two differentiable functions on an interval is differentiable
My attempt: I wanted to prove it from definition, writing
$$\lim_{h\to0}\frac{f_1(a+h)-f_1(a)}{h}+\lim_{h\to 0}\frac{f_2(a+h)-f_2(a)}{h}$$
Then I can write them under the same limit, but then I cannot proceed. Would be thankful if anyone helped me with it.
Best Answer
$$ \frac{f(x+h)-f(x)}{h}=\frac{f_1(x+h)+f_2(x+h)-f_1(x)-f_2(x)}{h}=\frac{f_1(x+h)-f_1(x)}{h}+\frac{f_2(x+h)-f_2(x)}{h}. $$ You are probably familiar with a standard theorem about limits that says that if $L:=\lim _{x\to a}f(x)$ exists and $M:=\lim _{x\to a}g(x)$ exists, then $\lim _{x\to a}[f+g](x)$ exists and is equal to $L+M$. Applying that result to the above equation tells us that the derivative of $f$ at $x$ exists and is equal to $f_1'(x)+f_2'(x)$.
If you need to prove the statement about limits, you'll have to do an $\epsilon$-$\delta$ argument.