Proving the derivative of a function of another function at a point Is the following equality true?
$$ \left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{x - x_0} $$
If it is true, how can I prove this?
It seems to work but I am unable to prove it. Take for example $ f(x) = x^2 $. Then $ f(ax + b) = (ax + b)^2 $.
\begin{align*} \text{LHS} &= \left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0} \\ \\
&= \left[ \frac{d}{dx} (ax + b)^2 \right]_{x = x_0} \\ \\
&= \left[ 2a(ax + b) \right]_{x = x_0} \\ \\
&= 2a(ax_0 + b). \\ \\
\text{RHS} &= \lim_{x \to x_0} \frac{(ax + b)^2 - (ax_0 + b)^2}{x - x_0} \\ \\
&= \lim_{x \to x_0} \frac{a^2x^2 + 2abx - a^2x_0^2 - abx_0}{x - x_0} \\ \\
&= \lim_{x \to x_0} \frac{a^2(x^2 - x_0^2) + 2ab(x - x_0)}{x - x_0} \\ \\
&= \lim_{x \to x_0} a^2(x + x_0) + 2ab \\ \\
&= 2a^2 x_0 + 2ab \\ \\
&= 2a(ax_0 + b).
\end{align*}
How do I prove this for any function $ f(x) $?

To do a formal proof, we can just use the chain rule.
\begin{align}
\left[ \frac{d}{dx} f(ax + b) \right]_{x = x_0}
&= \left[f'(x) \frac d{dx}(ax+b) \right]_{x=ax_0+b} \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{a(x-x_0)} a \\
&= \lim_{x \to x_0} \frac{f(ax + b) - f(ax_0 + b)}{(x-x_0)}
\end{align}
The first line is the chain derivative, the second line is the definition of the derivative.
