Let the function $f, g, h$ be a surjective function $\forall x \in R$, if $\frac{\text{d}}{\text{d}x}f(x) = g(x)$ and $h(x) = f(ax + b)$ where a, b is a constant and a $\ne$ 0.
Proof that $\frac{\text{d}}{\text{d}x}h(x) = ag(ax + b)$ using only the limits definition of derivative.
You can simply use the algebraic properties of limits.
Also, you can do it step by step: first step is with $b=0$: let $u(x):=f(ax)$, then $$\frac d{dx}u(x)=\lim_{h\to 0}\frac{f(a(x+h)) - f(ax)}h$$ Here observe that $h_2:=ah\to 0$, and use $\frac{f(ax+h_2)-f(ax)}{h_2}\to \frac d{dx}f(ax)$. The case $a=0$ can be separately handled.
The next step would be with $a=1$, so that $h(x)=u(x+b)$.
Then combine.