Derivatives of Function of a x + b

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

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

Then:
 * $\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$

where $z = a x + b$.

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$

where $z = a x + b$.

Basis for the Induction
$P(1)$ is the case:.
 * $\map {\dfrac \d {\d x} } {\map f {a x + b} } = a \map {\dfrac \d {\d z} } {\map f z}$

where $z = a x + b$.

This is proved in Derivative of Function of Constant Multiple: Corollary.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\map {\dfrac {\d^k} {\d x^k} } {\map f {a x + b} } = a^k \map {\dfrac {\d^n} {\d z^k} } {\map f z}$

where $z = a x + b$.

Then we need to show:
 * $\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\map f {a x + b} } = a^{k + 1} \map {\dfrac {\d^{k + 1} } {\d z^{k + 1} } } {\map f z}$

where $z = a x + b$.

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$

where $z = a x + b$.