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Using this definition:

If $f$ is a function and has derivative $f'(c)$ at the point $c$ in the domain of $f$ means that if ($a_n$)$_{n=1}^{\infty}$ is any sequence converging to $c$ such that $a_n$ $\not= c$is in the domain of $f$ for all $n \in \mathbb{N},$ then: $$\left[ \frac{f(x_n)-f(c)}{x_n-c}\right]_{n=1}^{\infty}$$converges to $f'(c)$

Assuming $f$ and $g$ are differentiable functions on (a,b) with $h(x)=f(x)+g(x),$ prove that $h'(x)=f'(x)+g'(x).$

Attempt so far:

Using the definition above, $\left[ \frac{h(x_n)-h(c)}{x_n-c}\right]_{n=1}^{\infty}$ converges to $h'(c)$ <=> $\left[ \frac{f(x_n)-f(c)}{x_n-c}\right]_{n=1}^{\infty}$ + $\left[ \frac{g(x_n)-g(c)}{x_n-c}\right]_{n=1}^{\infty}$ since $h(x)=f(x)+g(x).$

I'm not sure if using the fact that If ($a_n$) converges to $L$ and ($b_n$) converges to $K$ then ($a_n+b_n$) converges to $L+K$ would help at all? I feel like this proof should be somewhat simple looking at how it is proven using the normal derivative definition, but I am having trouble coming up with an actual proof.

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1 Answer 1

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I think you have got the essential elements already. If you still not convinced, I try to restate what you have said:

Pick any $c\in (a,b)$. Pick any sequence $(a_n)_{i= 1}^\infty$ such that $a_n \rightarrow c$ and $a_n \neq c$. We will show that $$\frac{h(x_n) - h(c)}{x_n - c} \rightarrow f'(c) + g'(c)$$

We assume that $f$ differentiable in $(a,b)$. So that $f'(c)$ and $g'(c)$ exists. From the def you provided, $$a_n=\frac{f(x_n) - f(c)}{x_n - c} \rightarrow f'(c) \text{ and } b_n=\frac{g(x_n) - g(c)}{x_n - c} \rightarrow g'(c) $$ The sequence $$\frac{h(x_n) - h(c)}{x_n - c} =\frac{[f(x_n)+g(x_n)] - [f(c)+g(c)]}{x_n - c} =a_n+b_n $$ so that it converges and the limit equals $f'(c) + g'(c)$. So that $h'(c) = f'(c)+g'(c)$ for any $c \in (a,b)$.

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