Sum Rule for Derivatives/General Result

Theorem
Let $\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x$ be real functions all differentiable.

Then for all $n \in \N_{>0}$:
 * $\ds \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$

Proof
The proof proceeds by induction.

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

$\map P 1$ is true, as this just says:
 * $\map {D_x} {\map {f_1} x} = \map {D_x} {\map {f_1} x}$

which is trivially true.

Basis for the Induction
$\map P 2$ is the case:
 * $\ds \map {D_x} {\map {f_1} x + \map {f_2} x} = \map {D_x} {\map {f_1} x} + \map {D_x} {\map {f_2} x}$

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 $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \map {D_x} {\sum_{i \mathop = 1}^k \map {f_i} x} = \sum_{i \mathop = 1}^k \map {D_x} {\map {f_i} x}$

from which it is to be shown that:
 * $\ds \map {D_x} {\sum_{i \mathop = 1}^{k + 1} \map {f_i} x} = \sum_{i \mathop = 1}^{k + 1} \map {D_x} {\map {f_i} x}$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \N_{>0}: \map {D_x} {\sum_{i \mathop = 1}^n \map {f_i} x} = \sum_{i \mathop = 1}^n \map {D_x} {\map {f_i} x}$