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| In calculus, the '''constant factor rule in differentiation''', also known as '''The Kutz Rule''', allows you to take [[Coefficient|constants]] outside a [[derivative]] and concentrate on [[derivative|differentiating]] the [[function (mathematics)|function]] of x itself. This is a part of the [[linearity of differentiation]].
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| Suppose you have a [[function (mathematics)|function]]
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| :<math>g(x) = k \cdot f(x).</math>
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| where ''k'' is a constant.
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| Use the formula for differentiation from first principles to obtain:
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| :<math>g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}</math>
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| :<math>g'(x) = \lim_{h \to 0} \frac{k \cdot f(x+h) - k \cdot f(x)}{h}</math>
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| :<math>g'(x) = \lim_{h \to 0} \frac{k(f(x+h) - f(x))}{h}</math> | |
| :<math>g'(x) = k \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \mbox{(*)}</math>
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| :<math>g'(x) = k \cdot f'(x).</math>
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| This is the statement of the constant factor rule in differentiation, in [[Lagrange's notation for differentiation]].
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| In [[Leibniz's notation]], this reads
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| :<math>\frac{d(k \cdot f(x))}{dx} = k \cdot \frac{d(f(x))}{dx}.</math>
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| If we put ''k''=-1 in the constant factor rule for differentiation, we have:
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| :<math>\frac{d(-y)}{dx} = -\frac{dy}{dx}.</math>
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| ==Comment on proof==
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| Note that for this statement to be true, ''k'' must be a [[Coefficient|constant]], or else the ''k'' can't be taken outside the [[limit of a function|limit]] in the line marked (*).
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| If ''k'' depends on ''x'', there is no reason to think ''k(x+h)'' = ''k(x)''. In that case the more complicated proof of the [[product rule]] applies.
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| [[Category:Differentiation rules]]
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