Integration by parts: Difference between revisions
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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]]. | |||
Suppose you have a [[function (mathematics)|function]] | |||
:<math>g(x) = k \cdot f(x).</math> | |||
where ''k'' is a constant. | |||
Use the formula for differentiation from first principles to obtain: | |||
:<math>g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}</math> | |||
:<math>g'(x) = \lim_{h \to 0} \frac{k \cdot f(x+h) - k \cdot f(x)}{h}</math> | |||
:<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> | |||
:<math>g'(x) = k \cdot f'(x).</math> | |||
This is the statement of the constant factor rule in differentiation, in [[Lagrange's notation for differentiation]]. | |||
In [[Leibniz's notation]], this reads | |||
:<math>\frac{d(k \cdot f(x))}{dx} = k \cdot \frac{d(f(x))}{dx}.</math> | |||
If we put ''k''=-1 in the constant factor rule for differentiation, we have: | |||
:<math>\frac{d(-y)}{dx} = -\frac{dy}{dx}.</math> | |||
==Comment on proof== | |||
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 (*). | |||
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. | |||
[[Category:Differentiation rules]] |
Revision as of 03:07, 22 January 2014
In calculus, the constant factor rule in differentiation, also known as The Kutz Rule, allows you to take constants outside a derivative and concentrate on differentiating the function of x itself. This is a part of the linearity of differentiation.
Suppose you have a function
where k is a constant.
Use the formula for differentiation from first principles to obtain:
This is the statement of the constant factor rule in differentiation, in Lagrange's notation for differentiation.
In Leibniz's notation, this reads
If we put k=-1 in the constant factor rule for differentiation, we have:
Comment on proof
Note that for this statement to be true, k must be a constant, or else the k can't be taken outside the limit in the line marked (*).
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.