Sum Rule for Derivatives/General Result

Theorem
Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be real functions all differentiable.

Then for all $n \in \N_{>0}$:
 * $\displaystyle D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$

Proof
The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$

$P \left({1}\right)$ is true, as this just says:
 * $D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$

which is trivially true.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\displaystyle D_x \left({f_1 \left({x}\right) + f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) + D_x \left({f_2 \left({x}\right)}\right)$

which has been proved in Sum Rule for Derivatives.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k D_x \left({f_i \left({x}\right)}\right)$

from which it is to be shown that:
 * $\displaystyle D_x \left({\sum_{i \mathop = 1}^{k + 1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k + 1} D_x \left({f_i \left({x}\right)}\right)$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N_{>0}: D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$