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:
 * $\dfrac {\rd^n} {\rd x^n} \left({f \left({a x + b}\right)}\right) = a^n \dfrac {\rd^n} {\rd z^n} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Proof
Proof by induction:

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

where $z = a x + b$.

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

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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\dfrac {\rd^k} {\rd x^k} \left({f \left({a x + b}\right)}\right) = a^k \dfrac {\rd^n} {\rd z^k} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Then we need to show:
 * $\dfrac {\rd^{k + 1} } {\rd x^{k + 1} } \left({f \left({a x + b}\right)}\right) = a^{k + 1} \dfrac {\rd^{k + 1} } {\rd z^{k + 1} } \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Induction Step
This is our induction step:

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

Therefore:
 * $\dfrac {\rd^n} {\rd x^n} \left({f \left({a x + b}\right)}\right) = a^n \dfrac {\rd^n} {\rd z^n} \left({f \left({z}\right)}\right)$

where $z = a x + b$.