My book says that $\left(x^r\right)'=rx^{r-1}$ so $\left(4x\right)'=1\cdot 4^{1-1}=4^{0}$, but $a^{0}=1$?
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1The $4$ is a multiplier, and remember that differentiation is a linear operation. In other words, for any function, $$ (4f)' = 4f' $$ And in this case, $f = x$, $f' = 1$ so $(4x)' = 4$. – Matti P. Apr 26 '18 at 07:07
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2so, $(4x)'=(4\cdot1x^{1-1})=4$ – Just dropped in Apr 26 '18 at 07:07
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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
$$(4\cdot x)'=4\cdot(1\cdot x^{1-1})=4.$$
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1I imagine the deeper error here is not just misapplying the rule but misunderstanding the role of "$x$" as the input to a function being differentiated, rather than a variable that can stand for any number. – Eric Wofsey Apr 26 '18 at 07:17
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By applying the chain rule, one has $$ \left[\,f(ax)\,\right]'=a\cdot f'(ax)\color{red}{\ne} f'(ax). $$
Olivier Oloa
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