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

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. – Arturo Magidin May 23 '12 at 3:13
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 $x$? – Rahul May 23 '12 at 3:16
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). – anon May 23 '12 at 3:16
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! – ninja08 May 23 '12 at 3:18
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. – Arturo Magidin May 23 '12 at 3:26

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: $$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \to 0}\dfrac{x+h-x}{h} = \lim_{h \to 0} \; 1 = 1.$$
lim $0 f(x)$ is always $0$. – Gastón Burrull May 23 '12 at 3:27
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. – Eugene May 23 '12 at 3:28
Because it is simply the constant function. So $f(x) = c$ for all $x$. Therefore $f(x+h) = c$ and $f(x) = c$. – Eugene May 23 '12 at 14:31
$f(x)=x^1, f'(x)=1(x)^{1-1}=1x^0=1$ I simplified this problem as much as I could. I hope this helped. And also the power rule states $f(x)=x^n$ and $f'(x)=n(x)^{n-1}$.