Sum Rule for Derivatives
Contents |
Theorem
Let $f \left({x}\right), j \left({x}\right), k \left({x}\right)$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $f \left({x}\right) = j \left({x}\right) + k \left({x}\right)$.
Then:
- $f^{\prime} \left({\xi}\right) = j^{\prime} \left({\xi}\right) + k^{\prime} \left({\xi}\right)$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
- $\forall x \in I: f^{\prime} \left({x}\right) = j^{\prime} \left({x}\right) + k^{\prime} \left({x}\right)$
General Result
Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be real functions all differentiable as above.
Then:
- $\displaystyle D_x \left({\sum_{i=1}^n f_i \left({x}\right)}\right) = \sum_{i=1}^n D_x \left({f_i \left({x}\right)}\right)$
Proof 1
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle f^{\prime} \left({\xi}\right)\) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {f \left({\xi+h}\right) - f \left({\xi}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by the definition of the derivative | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {\left({j \left({\xi+h}\right) + k \left({\xi+h}\right)}\right) - \left({j \left({\xi}\right) + k \left({\xi}\right)}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {j \left({\xi+h}\right) + k \left({\xi+h}\right) - j \left({\xi}\right) - k \left({\xi}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {\left({j \left({\xi+h}\right) - j \left({\xi}\right)}\right) + \left({k \left({\xi+h}\right) - k \left({\xi}\right)}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \left({\frac {j \left({\xi+h}\right) - j \left({\xi}\right)} h + \frac {k \left({\xi+h}\right) - k \left({\xi}\right)} h}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {j \left({\xi+h}\right) - j \left({\xi}\right)} h + \lim_{h \to 0} \frac {k \left({\xi+h}\right) - k \left({\xi}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle j^{\prime} \left({x}\right) + k^{\prime} \left({x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by the definition of the derivative |
$\blacksquare$
Proof 2
It can be observed that this is an example of a Linear Combination of Derivatives with $\lambda = \mu = 1$.
$\blacksquare$
Proof of General Result
Follows directly by induction.
$\blacksquare$
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