constant factor rule in differentiation

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

g(x) = k \cdot f(x).

Use the formula for differentiation from first principles to obtain:

g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}

g'(x) = \lim_{h \to 0} \frac{k \cdot f(x+h) - k \cdot f(x)}{h}

g'(x) = \lim_{h \to 0} \frac{k(f(x+h) - f(x))}{h}

g'(x) = k \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \mbox{(*)}

g'(x) = k \cdot f'(x).

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

\frac{d(k \cdot f(x))}{dx} = k \cdot \frac{d(f(x))}{dx}.

If we put k=-1 in the constant factor rule for differentiation, we have:

\frac{d(-y)}{dx} = -\frac{dy}{dx}.

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.

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