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

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:

$$ $$ $$ $$ $$

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 $$z = a x + b$$.