# Derivative of Constant Multiple/Real/Corollary

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## Corollary to Derivative of Constant Multiple

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

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

Then:

- $\map {\dfrac {\d^n} {\d x^n} } {c \map f x} = c \map {\dfrac {\d^n} {\d x^n} } {\map f x}$

## Proof

By induction: the base case is for $n = 1$ and is proved in Derivative of Constant Multiple.

Now consider $\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {c \map f x}$, assuming the induction hypothesis $\map {\dfrac {\d^k} {\d x^k} } {c \map f x} = c \map {\dfrac {\d^k} {\d x^k} } {\map f x}$:

\(\displaystyle \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {c \map f x}\) | \(=\) | \(\displaystyle \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d x^k} } {c \map f x} }\) | Definition of Higher Derivative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\dfrac \d {\d x} } {c \map {\dfrac {\d^k} {\d x^k} } {\map f x} }\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle c \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d x^k} } {\map f x} }\) | Basis for the Induction | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle c \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\map f x}\) | Definition of Higher Derivative |

Hence the result by induction.

$\blacksquare$