Derivative of a constant not making sense I'm studying calculus on my own and I came across something weird:
The derivative of $x^n$ is $n\cdot x^{n-1}$, and the derivative of any constant is $0$. Also, any constant $x = x^1$. However, the derivative of $x^1$ ($x$ being any integer constant) seems to be $1$: $n\cdot x^{n-1} = 1\cdot x^{1-1} = x^0 = 1$. What's failing in my reasoning?
 A: Well, first off 
$f(x) = x = x^1$ is not a constant function by any stretch of the imagination.
You then said to replace it with a constant so
Let $f(9) = 9 = 9^1$.  Now, technically this is a function but it is a function that maps from a space that has only one point.  It is not a function that has any variation at all.  It exists at one point only.
The derivative $f'(x) = \lim_{x+\delta x \rightarrow x} \frac{f(x + \delta x) - f(x)}{\delta x}$ which measures rate of variation just doesn't make any sense when $x$ never varies. When $x = 9$ for all $x$ always; there is no other $x$ in existence and $\delta 9 = 0$ always because $9$ never varies-- it is always 9, such an expression makes no sense.
It would have to equal $f'(9) = \frac{f(9 + \delta 9) - f(9)}{\delta 9} = (9-9)/0= 0/0$ which is undefinable.
Now you note that $f(x) = x^n$ then $f'(x) = n*x^{n-1}$.  So if you have $f(x) = 9 = 9*x^0$ one might suppose $f'(x) = 9*0*x^{n-1} = 0$.  That would almost be right.
$dx^n/dx = \lim_{h\rightarrow 0} \frac{(x+h)^n - x^n}{h} = \lim \frac{x^n + h*x^{n-1} + {n \choose 2}h^2x^{n-2} +.... + h^n - x^n}{h} = \lim x^n/h + hn*x^{n-1} + {n \choose 2}h*x^{n-2}+... - h^{n-1} - x^{n}/h= \lim hn*x^{n-1}/h = nx^{n-1}$.  But this assumes $n > 0$.  If $n = 0$ we get $dx^n/dx = \lim_{h\rightarrow 0} \frac{(x+h)^n - x^n}{h}= \lim \frac{(x+h)^0 - x^0}{h} = \lim \frac 0 h = 0$.
No matte how you look at it: if $f(x) = k$ then $f'(k) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h} = \lim \frac{k-k}{h} = \lim \frac 0h = 0$.
A: Let $a$ be some number and $D$ be the differential operator (the derivative).  Just for fun we'll use the notation $x\mapsto f(x)$ to represent the function $f$.  This notation is supposed to be pretty intuitive: in it $x$ is an arbitrary element in the domain of $f$ and $f(x)$ is the number it maps to under $f$.
We know that $$D(x\mapsto x^n) = x\mapsto nx^{n-1}\tag{1}$$ and that $$D(x\mapsto a) = x\mapsto 0\tag{2}$$
So then does that mean that $D(x\mapsto a^n) \stackrel{?}= x\mapsto na^{n-1}$?  No.  Of course not.  $a^n$ is just another number so we follow rule $(2)$ to get $$D(x\mapsto a^n) = D(x\mapsto b) = x\mapsto 0$$ where $b=a^n$.
A: I don't think you understand the definition of a derivative. Go back to the definition of a derivative, and review the first principle. You are trying to take the instantaneous rate of change of any function. There is a clear difference between $f(x) = x$ and $f(x) = c$, $c$ $\epsilon$ $\mathbb{R}$. The power rule, $\frac{d}{dx}x^n = nx^{n-1}$ only applies to functions where $x$ is the variable. When the function is a constant, that means that the rate of change is 0. In your example, you are using $x$ as a placeholder for a constant value, and that is functionally different from using $x$ as a variable.
The next time you have a question about derivatives, you can always make sure if it makes sense with the first principle of derivatives:
$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
A lot of people devalue this theorem, but it's extremely fundamental and important in calculus proofs.
If the degree of your function is low, you can also think of functions graphically and try to picture it's rate of change. In this case, if you thought of the difference between $f(x) = x$ and $f(x) = c$ (in your case you mistakenly used x), then you'd see the flaw in your reasoning.
A: Please note that a constant is not a  constant function. We can try to derive functions, we can surely derive constant functions, but we certainly do not derive constants.  So, yes, anyway, derivative of a constant doesn't make any sense.
Another preliminary remark : the expression $x^n$ in the function $f : x \mapsto x^n $ is the value of the function on $x$, not the function. If you write the function $x^n$ (or worse $x^n$ alone) you should keep in mind that it's just a (bad) shorthand notation. For example you can't really derive $x^n$ and $c$ you derive the functions $f : x \mapsto x^n $  and $g : x \mapsto c$. If you forget that it can lead you to mistakes...
What's wrong in your reasoning is that you used the $x$ symbol with $2$ different meanings without noticing it.
When you write $x=x^1$ and say $x$ is a constant it's ok, it could be, even if we generally use $x$ as a variable name. But when you say you derive $x^1$ and its derivative is $1$, implicitly, you derive the identity function ($Id : x \mapsto x $) where $x$ represents no longer constant. Try to use different names for your constants and your variables in function definitions and it should be ok.  For example if you want a constant function whose value is $x$ write $f : t \mapsto x $ and if you don't really need to name $x$ the constant you could use this more common form : $f : x \mapsto c $ where $c$ is the constant and $x$ the variable. 
