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:
\(\ds \map {D_x} {\sum_{i \mathop = 1}^{k + 1} \map {f_i} x}\) | \(=\) | \(\ds \map {D_x} {\sum_{i \mathop = 1}^k \map {f_i} x + \map {f_{k + 1} } x}\) | 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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: General Rules of Differentiation: $13.5$