Sum Rule for Derivatives/General Result

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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:


\(\ds \map {D_x} {\sum_{i \mathop = 1}^{k + 1} \map {f_i} x}\) \(=\) \(\ds \map {D_x} {\sum_{i \mathop = 1}^k \map {f_i} x + \map {f_{k + 1} } x}\) Definition of Summation
\(\ds \) \(=\) \(\ds \map {D_x} {\sum_{i \mathop = 1}^k \map {f_i} x} + \map {D_x} {\map {f_{k + 1} } x}\) Basis of the Induction
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k \map {D_x} {\map {f_i} x} + \map {D_x} {\map {f_{k + 1} } x}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{k + 1} \map {D_x} {\map {f_i} x}\) Definition of Summation

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


Therefore:

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

$\blacksquare$


Sources

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