# 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:

- $\map {D_x} {\map f {a x + b} } = a \, \map {D_{a x + b} } {\map f {a x + b} }$

## 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 for Derivatives | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle c \map {D_{c x} } {\map f {c x} }\) | from above |

$\blacksquare$