Derivatives of Function of ax + b

Theorem

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

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

Then:

$D^n_x \left({f \left({a x + b}\right)}\right) = a^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$D^n_x \left({f \left({a x + b}\right)}\right) = a^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Basis for the Induction

$P(1)$ is the case:.

$D_x \left({f \left({a x + b}\right)}\right) = a D_{z} \left({f \left({z}\right)}\right)$

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:

$D^k_x \left({f \left({a x + b}\right)}\right) = a^k D^k_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Then we need to show:

$D^{k+1}_x \left({f \left({a x + b}\right)}\right) = a^{k+1} D^{k+1}_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D^{k+1}_x \left({f \left({a x + b}\right)}\right)\)	\(=\)	\(\displaystyle D_x \left({D^k_x \left({f \left({a x + b}\right)}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of higher derivatives
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle D_x \left({a^k D^k_{z} \left({f \left({z}\right)}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^k D_x \left({D^k_{ax + b} \left({f \left({ax + b}\right)}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Constant Multiple
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^k \cdot a D_z \left({D^k_{z} \left({f \left({z}\right)}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the base case
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^{k+1} D_z \left({D^k_{z} \left({f \left({z}\right)}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

Therefore:

$D^n_x \left({f \left({a x + b}\right)}\right) = a^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

$\blacksquare$