Constant factor rule in differentiation

Suppose you have a function

<math>g(x) = k \cdot f(x)</math>

Differentiate both sides to obtain:

<math>g'(x) = \frac{d(k \cdot f(x))}{dx}</math>

where the notation on the LHS is Newton's notation for differentiation and on the RHS we have Leibniz's notation for differentiation.

Now 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 g(x)}{h}</math>

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

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

<math>g'(x) = k \cdot f'(x)</math>

But we have already seen that:

<math>g'(x) = \frac{d(k\cdot f(x))}{dx}</math>

Thus:

<math>\frac{d(k \cdot f(x))}{dx} = k \cdot f'(x)</math>

Or:

<math>\frac{d(k \cdot f(x))}{dx} = k \cdot \frac{d(f(x))}{dx}</math>

This is the statement of the constant factor rule in differentiation.

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 (*). The reason for this taking away follows from the product rule for limits[?]:

<math>\lim_{x \to 0} kf(x) = (\lim_{x \to 0} k) \cdot (\lim_{x \to 0} f(x)) = k \lim_{x \to 0} f(x)</math>

(since x does not affect k).

Hence you CAN take the k outside the limit, since k is a constant with respect to x, and so isn't going to make any difference to k if x tends to 0.

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

<math>\frac{d(-y)}{dx} = -\frac{dy}{dx}</math>