Why is $(4x)' \neq 1$?

Question

My book says that $\left(x^r\right)'=rx^{r-1}$ so $\left(4x\right)'=1\cdot 4^{1-1}=4^{0}$, but $a^{0}=1$?

Answer 1

You are applying the rule in a weird (and wrong) way.

Comparing $4x$ and $x^r$, you have that $r=1$, and there is an extra factor $4.$ So

$$(4\cdot x)'=4\cdot(1\cdot x^{1-1})=4.$$

Answer 2

By applying the chain rule, one has $$ \left[\,f(ax)\,\right]'=a\cdot f'(ax)\color{red}{\ne} f'(ax). $$