Derivative of Constant Multiple/Real/Corollary

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

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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 $\map {\dfrac {\d^k} {\d x^k} } {c \map f x} = c \map {\dfrac {\d^k} {\d x^k} } {\map f x}$:

\(\ds \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {c \map f x}\)

\(=\)

\(\ds \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d x^k} } {c \map f x} }\)

Definition of Higher Derivative

\(\ds \)

\(=\)

\(\ds \map {\dfrac \d {\d x} } {c \map {\dfrac {\d^k} {\d x^k} } {\map f x} }\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds c \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d x^k} } {\map f x} }\)

Basis for the Induction

\(\ds \)

\(=\)

\(\ds c \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\map f x}\)

Definition of Higher Derivative

Hence the result by induction.

$\blacksquare$