Derivative of Function of Constant Multiple

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Theorem

Let $f$ be a real function which is differentiable on $\R$.

Let $c \in \R$ be a constant.


Then:

$\map {D_x} {\map f {c x} } = c \, \map {D_{c x} } {\map f {c x} }$


Corollary

Let $a, b \in \R$ be constants.


Then:

$D_x \left({f \left({a x + b}\right)}\right) = a D_{a x + b} \left({f \left({a x + b}\right)}\right)$


Proof

First it is shown that $\map {D_x} {c x} = c$:

\(\displaystyle \map {D_x} {c x}\) \(=\) \(\displaystyle c \, \map {D_x} x + x \, \map {D_x} c\) Product Rule
\(\displaystyle \) \(=\) \(\displaystyle c + x \, \map {D_x} c\) Derivative of Identity Function
\(\displaystyle \) \(=\) \(\displaystyle c + 0\) Derivative of Constant
\(\displaystyle \) \(=\) \(\displaystyle c\)


Next:

\(\displaystyle \map {D_x} {\map f {c x} }\) \(=\) \(\displaystyle \map {D_x} {c x} \, \map {D_{c x} } {\map f {c x} }\) Chain Rule
\(\displaystyle \) \(=\) \(\displaystyle c \, \map {D_{c x} } {\map f {c x} }\) from above

$\blacksquare$