Constant factor rule in differentiation

In calculus, the constant factor rule in differentiation, also known as the Kutz Rule , 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

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

where k is a constant.

Use the formula for differentiation from first principles to obtain:

$g'(x)=\lim _{{h\to 0}}{\frac {g(x+h)-g(x)}{h}}$

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

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

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

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

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

In Leibniz's notation, this reads

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

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.