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			{{Calculus \|Differential}}

			In [[calculus]], the '''sum rule in differentiation''' is a method of finding the [[derivative]] of a function that is the sum of two other functions for which derivatives exist. This is a part of the [[linearity of differentiation]]. The [[sum rule in integration]] follows from it. The rule itself is a direct consequence of differentiation from [[first principles]].

			The sum rule tells us that for two [[function (mathematics)\|function]]s ''u'' and ''v'':

			:<math>\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}</math>

			This rule also applies to subtraction and to additions and subtractions of more than two functions
			:<math>\frac{d}{dx}(u + v + w + \dots)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\cdots</math>

			==Proof==
			Let ''y'' be a [[function (mathematics)\|function]] given by the [[sum]] of two [[function (mathematics)\|functions]] ''u'' and ''v'', such that:

			:<math> y = u + v \,</math>

			Now let ''y'', ''u'' and ''v'' be increased by small increases Δ''y'', Δ''u'' and Δ''v'' respectively. Hence:

			:<math> y + \Delta{y} = (u + \Delta{u}) + (v + \Delta{v}) = u + v + \Delta{u} + \Delta{v} = y + \Delta{u} + \Delta{v}. \, </math>

			So:

			:<math> \Delta{y} = \Delta{u} + \Delta{v}. \, </math>

			Now divide throughout by Δ''x'':

			:<math> \frac{\Delta{y}}{\Delta{x}} = \frac{\Delta{u}}{\Delta{x}} + \frac{\Delta{v}}{\Delta{x}}. </math>

			Let Δ''x'' tend to 0:

			:<math> \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}. </math>

			Now recall that ''y'' = ''u'' + ''v'', giving the sum rule in differentiation:

			:<math> \frac{d}{dx}\left(u + v\right) = \frac{du}{dx} + \frac{dv}{dx} .</math>

			The rule can be extended to [[subtraction]], as follows:

			:<math> \frac{d}{dx}\left(u - v\right) = \frac{d}{dx}\left(u + (-v)\right) = \frac{du}{dx} + \frac{d}{dx}\left(-v\right). </math>

			Now use the special case of the [[constant factor rule in differentiation]] with ''k''=−1 to obtain:

			:<math> \frac{d}{dx}\left(u - v\right) = \frac{du}{dx} + \left(-\frac{dv}{dx}\right) = \frac{du}{dx} - \frac{dv}{dx}. </math>

			Therefore, the sum rule can be extended so it "accepts" [[addition]] and [[subtraction]] as follows:

			:<math> \frac{d}{dx}\left(u \pm v\right) = \frac{du}{dx} \pm \frac{dv}{dx}. </math>

			The sum rule in differentiation can be used as part of the [[derivation (calculus)\|derivation]] for both the [[sum rule in integration]] and [[linearity of differentiation]].

			==Generalization to finite sums==
			Consider a set of functions ''f''<sub>1</sub>, ''f''<sub>2</sub>,..., ''f''<sub>''n''</sub>. Then

			:<math> \frac{d}{dx} \left(\sum_{1 \le i \le n} f_i(x)\right) = \frac{d}{dx}\left(f_1(x) + f_2(x) + \cdots + f_n(x)\right) = \frac{d}{dx}f_1(x) + \frac{d}{dx}f_2(x) + \cdots + \frac{d}{dx}f_n(x) </math>
			so
			:<math> \frac{d}{dx} \left(\sum_{1 \le i \le n} f_i(x)\right) = \sum_{1 \le i \le n} \left(\frac{d}{dx}f_i(x)\right) .</math>

			In other words, the derivative of any finite [[sum]] of functions is the sum of the derivatives of those functions.

			This follows easily by induction; we have just proven this to be true for ''n'' = 2. Assume it is true for all ''n'' < ''k'', then define

			:<math>g(x)=\sum_{i=1}^{k-1} f_i(x).</math>

			Then

			:<math>\sum_{i=1}^k f_i(x)=g(x)+f_k(x)</math>

			and it follows from the proof above that

			:<math> \frac{d}{dx} \left(\sum_{i=1}^k f_i(x)\right) = \frac{d}{dx}g(x)+\frac{d}{dx}f_k(x).</math>

			By the inductive hypothesis,

			:<math>\frac{d}{dx}g(x)=\frac{d}{dx} \left(\sum_{i=1}^{k-1} f_i(x)\right)=\sum_{i=1}^{k-1} \frac{d}{dx}f_i(x)</math>

			so

			:<math>\frac{d}{dx} \left(\sum_{i=1}^k f_i(x) \right) = \sum_{i=1}^{k-1} \frac{d}{dx}f_i(x) + \frac{d}{dx}f_k(x)=\sum_{i=1}^k \frac{d}{dx}f_i(x) </math>

			which ends the proof of the sum rule of differentiation.

			Note this does not automatically extend to infinite sums. An intuitive reason for why things can go wrong is that there is more than one limit involved (specifically, one for the sum and one in the definition of the [[derivative]]). [[Uniform convergence]] deals with these sorts of issues.

			==References==
			*Gilbert Strang: ''Calculus''. SIAM 1991, ISBN 0-9614088-2-0, p. 71 ([http://books.google.de/books?id=OisInC1zvEMC&pg=PA197&dq=%22sum+rule%22+calculus&as_brr=3&sig=ACfU3U0B8K2WU1A1vbYE605MWxSV8YwAKA restricted online version (google books)])
			*[http://planetmath.org/encyclopedia/SumRule.html sum rule at [[PlanetMath]]]

			[[Category:Differentiation rules]]

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