# Constant factor rule in differentiation

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In calculus, the constant factor rule in differentiation allows one to take constants outside a derivative and concentrate on differentiating the function of x itself. This is a part of the linearity of differentiation.

Consider a differentiable function

${\displaystyle g(x)=k\cdot f(x).}$

where k is a constant.

Use the formula for differentiation from first principles to obtain:

${\displaystyle g'(x)=\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}}$
${\displaystyle g'(x)=\lim _{h\to 0}{\frac {k\cdot f(x+h)-k\cdot f(x)}{h}}}$
${\displaystyle g'(x)=\lim _{h\to 0}{\frac {k(f(x+h)-f(x))}{h}}}$
${\displaystyle g'(x)=k\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$
${\displaystyle g'(x)=k\cdot f'(x).}$

This is the statement of the constant factor rule in differentiation, in Lagrange's notation for differentiation.

In Leibniz's notation, this reads

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

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

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

## 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.