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:
| \(\ds \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\map f {a x + b} }\) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d x^k} } {\map f {a x + b} } }\) | Definition of Higher Derivative | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {a^k \map {\dfrac {\d^k} {\d z^k} } {\map f z} }\) | Induction Hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^k \map {\dfrac \d {\d x} } {\map {\dfrac {\d^k} {\d z^k} } {\map f z} }\) | Derivative of Constant Multiple | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^k \cdot a \map {\dfrac \d {\d z} } {\map {\dfrac {\d^k} {\d z^k} } {\map f z} }\) | Basis of the Induction | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^{k + 1} \map {\dfrac \d {\d z} } {\map {\dfrac {\d^k} {\d z^k} } {\map f z} }\) |
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 $z = a x + b$.
$\blacksquare$