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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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    Well, for one thing, the derivative is the slope of the tangent to the graph. A number represents a constant function, which is a straight *horizontal* line, with slope $0$, whereas $x$ represents the function $y=x$, which is a straight line with slope $1$. As to your second sentence, I don't understand what "it changes for number a variable for a number"; I can't even parse it.2012-05-23
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    The derivative of something with respect to $x$ is how much it changes when you change $x$. How much does the number 4 change when you change $x$?2012-05-23
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    You need to think of them as *functions*; one is a constant function (hence zero rate-of-change) while the other function increases linearly (hence positive constant rate-of-change).2012-05-23
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    You've explained it perfectly for me. I forgot that the derivative is a slope. The 2nd sentence was just trying to say why a variable, or placeholder for a number like 'x', would be equal to 1 while a number is equal to 0. But now I understand it's because a derivative is a slope. Thanks!2012-05-23
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    Constant functions doesn't change, then derivative is 0. $f(x)=x$ change if you evaluate in different points.2012-05-23
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    In the context of functions and especially of calculus, variables are not merely "placeholders for numbers". If you are thinking about $x$ as some fixed, but unknown, number then, to quote the Sage of Dagobah, "you must *unlearn* what you have learned." , you are *not* going to understand *anything* that is going on. You need to shake that preconception.2012-05-23
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    Ahh ok, I see Gaston. I wish I could give points or rep for these comments since they are helping a lot. Arturo - So what your trying to say is that x IS a number? I think Gaston was trying to say that by explaining that x is a number that changes. Thanks everyone.2012-05-23
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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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    lim $0 f(x)$ is always $0$.2012-05-23
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    You're right. I was overthinking this as I was worrying if I needed to consider $lim \frac{0}{0}$. This doesn't happen in this case though so you're right.2012-05-23
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    You can't go wrong by using the definition.2012-05-23
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    The derivative of x is 1 if you are taking the derivative wrt x. If you take it wrt another variable (say x's good friend y), the derivative os zero.2012-05-23
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    In the first example with the constants, how did you get c-c? Thank you for the reply btw.2012-05-23
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    Because it is simply the constant function. So $f(x) = c$ for all $x$. Therefore $f(x+h) = c$ and $f(x) = c$.2012-05-23
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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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