Constant factor rule in differentiation
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In calculus, the constant factor rule in differentiation allows you to take constants outside a derivative and concentrate on differentiating the function of x itself.
Suppose you have a function
- <math>g(x) = k \cdot f(x).<math>
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 for differentiation, 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 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.