Derivatives of Function of a x + b

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Theorem

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

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


Then:

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

where $z = a x + b$.


Proof

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

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

where $z = a x + b$.


Basis for the Induction

$P(1)$ is the case:.

$\dfrac \d {\d x} \left({f \left({a x + b}\right)}\right) = a \dfrac \d {\d z} \left({f \left({z}\right)}\right)$

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:

$\dfrac {\d^k} {\d x^k} \left({f \left({a x + b}\right)}\right) = a^k \dfrac {\d^n} {\d z^k} \left({f \left({z}\right)}\right)$

where $z = a x + b$.


Then we need to show:

$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \left({f \left({a x + b}\right)}\right) = a^{k + 1} \dfrac {\d^{k + 1} } {\d z^{k + 1} } \left({f \left({z}\right)}\right)$

where $z = a x + b$.


Induction Step

This is our induction step:

\(\displaystyle \dfrac {\d^{k+1} } {\d x^{k+1} } \left({f \left({a x + b}\right)}\right)\) \(=\) \(\displaystyle \dfrac \d {\d x} \left({\dfrac {\d^k} {\d x^k} \left({f \left({a x + b}\right)}\right)}\right)\) Definition of Higher Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \dfrac \d {\d x} \left({a^k \dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle a^k \dfrac \d {\d x} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)\) Derivative of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle a^k \cdot a \dfrac \d {\d z} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)\) Basis of the Induction
\(\displaystyle \) \(=\) \(\displaystyle a^{k+1} \dfrac \d {\d z} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)\)

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


Therefore:

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

where $z = a x + b$.

$\blacksquare$