# Proving the sum of two differentiable functions is also differentiable

I have to prove that

If $$f$$ and $$g$$ are differentiable at some point $$c$$, i.e. $$f'(c)$$ and $$g'(c)$$ exist, then $$(f+g)(x)$$ is also differentiable, i.e. $$(f+g)'(x)$$ exists.

In all the answers given everywhere, people have shown just that $$(f+g)'(x)=f'(x)+g'(x)$$. But that doesn't really show that the function is differentiable in a rigorous way (Does it?). I mean that shouldn't we use the $$\epsilon-\delta$$ to show actually that the limit exists?

I tried to show that given

$$\begin{gather} 0 < \lvert x-c \rvert < \delta_1 \implies \Biggl\lvert\frac{f(x)-f(c)}{x-c} - f'(c)\Biggr\rvert < \epsilon_1 \\ 0 < \lvert x-c \rvert < \delta_2 \implies \Biggl\lvert\frac{g(x)-g(c)}{x-c} - g'(c)\Biggr\rvert < \epsilon_2 \end{gather}$$

we somehow need to show that if we set

$$\delta = \min\{\delta_1, \delta_2\}$$

then

$$\forall \lvert x - \delta \rvert < 0, \qquad \Biggl\lvert\frac{(f+g)(x)-(f+g)(c)}{x-c}-L\Biggr\rvert < \epsilon$$

But I can't proceed to prove this.

• You don't need to reinvent the wheel. It is not necessary to use the $\epsilon$-$\delta$ language in order to be rigorous! One can apply proved lemmas/theorems/properties one needs.
– user9464
Commented Oct 26, 2019 at 19:58
• What you want to prove is $f+g$ is differentiable at $x=c$ and $(f+g)'(c)=f'(c)+g'(c)$. Otherwise, your statement is not necessarily true.
– user9464
Commented Oct 26, 2019 at 20:00
• @Jack Yes that is what I want Commented Oct 26, 2019 at 20:03
• This is essentially $\lim_{x\to c}(F(x)+G(x))=\lim_{x\to c}F(x)+\lim_{x\to c}G(x)$ when the limits on the right exist.
– user9464
Commented Oct 26, 2019 at 20:05
• @Jack Yes I have proved that for limits Commented Oct 26, 2019 at 20:07

It's easier to show that if you notice that $$L = f'(c) + g'(c)$$, then $$\begin{eqnarray} \left|\frac{(f+g)(x) - (f+g)(c)}{x - c} - (f'(c) + g'(c))\right| &\leq& \left| \frac{f(x) - f(c)}{x - c} - f'(c) \right| + \left| \frac{g(x) - g(c)}{x - c} - g'(c) \right| \\&<& \epsilon_{1} + \epsilon_{2}. \end{eqnarray}$$

• Yeah, that's right. But the question goes like this- "Prove that $(f+g)'(x)$ is differentiable and is equal to $f'(x)+g'(x)$".. so while proving differentiability at first we shouldn't assume the result of L right? Commented Oct 26, 2019 at 20:00
• There is no problem in guessing a value for $L$ because of the uniqueness of the limit.
– ABP
Commented Oct 26, 2019 at 20:03
• Oh yes. That makes sense :-) Commented Oct 26, 2019 at 20:03

$$0<|x-c|<\delta$$ implies $$|\frac{f(x)+g(x)-f(c)-g(c)}{x-c}-(f'(c)+g'(c))|$$ $$\le |\frac{f(x)-f(c)}{x-c}-f'(c)|+|\frac{g(x)-g(c)}{x-c}-g'(c)|$$ by the triangle inequality.

If $$f(x)$$ is differentiable at $$c$$ then by definition, $$f'(c)=\lim\limits_{h\to 0}\dfrac{f(c+h)-f(c)}{h}.$$ Similarly, since $$g(x)$$ is differentiable at $$c,$$ then $$g'(c) = \lim\limits_{h\to 0}\dfrac{g(c+h)-g(c)}{h}.$$ By definition, $$(f + g)(x)=f(x)+g(x).$$ Now we want to show that $$\lim\limits_{h\to 0}\dfrac{[f(c+h)+g(c+h)]-[f(c)+g(c)]}{h}=f'(c)+g'(c).$$ Simply manipulate the terms and use limit properties to get

$$\lim\limits_{h\to 0}\dfrac{[f(c+h)+g(c+h)]-[f(c)+g(c)]}{h}=\lim\limits_{h\to 0}\dfrac{f(c+h)-f(c)}{h} -\lim\limits_{h\to 0}\dfrac{g(c+h)-g(c)}{h}$$ $$=f'(c)+g'(c)$$ by the above definition, as required.

We can also use that since $$f$$ and $$g$$ are differentiable at $$x=c$$ that is

• $$f(c+h)=f(c)+h\cdot f'(c)+o(h)$$

• $$g(c+h)=g(c)+h\cdot g'(c)+o(h)$$

then we have

$$\begin{split} (f+g)(c+h) &= f(c+h)+g(c+h) \\ &= f(c)+g(c)+h\cdot(f'(c)+ g'(c))+o(h) \\ &= (f+g)(c)+h\cdot (f+g)'(c)+o(h) \end{split}$$

that is also $$(f+g)$$ is differentiable at $$x=c$$.