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why is the derivative of a number 0 while the derivative of $x$ is 1?

I can't understand why it changes for number and a variable for a number.

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    understand what is a function FIRST. only after that (2 years later) look at derivatives2016-02-06

3 Answers 3

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For a constant, let $f(x) = c$, where $c$ is a constant. Then we have that by the definition of a derivative that: $ f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \; \dfrac{c - c}{h} = \lim_{h \to 0} \; \dfrac{0}{h} = \lim_{h \to 0} \; 0 = 0 $ and for $f(x) = x$ we have that: $ f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\dfrac{x+h-x}{h} = \lim_{h \to 0} \; 1 = 1. $

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    ah ok I get that now.2012-05-23
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$f(x)=x^1, f'(x)=1(x)^{1-1}=1x^0=1$ I simplified this problem as much as I could. I hope this helped. And also the power rule states $f(x)=x^n$ and $f'(x)=n(x)^{n-1}$.

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If in doubt, always plot the graph. It becomes clear why the derivative of constant functions are zero - clearly as $x$ changes, $y$ doesn't. You can also see how this differs from $y(x) = x$.

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