Why is 1 the derivative of $x$? Can someone explain to me why the derivative of $x$ is 1. We just started derivatives in my calculus class and I don't understand it at all.
 A: The derivative tells us the slope of a line. The function $y(x)=x$ is a line with slope $1$. So if the derivative is to make any sense and actually tell us the slope, the derivative of $x$ should be $1$.
In general, for functions that look like $x^n$, their derivatives are $nx^{n-1}$, i.e. move the power out in front and reduce the power by $1$.
A: Draw the graph of $y = x$; it is a line with slope 1. If $f$ is a function, then $f'(a)$ is the slope of the line tangent to $f$'s graph at $a$.  In this case, the only way a line can be tangent to a line is if it is the line itself.  Hence 
if $f(x)= x$,
$f'(x) = 1$ for any $x$.
A: There's this formula to find derivatives of monomials, like $x^3$ and $5x^2$. What you do is you multiply the term by its power, then subtract the power by $1$. 
so, $x^2$ becomes $2\cdot x^{2-1} = 2x$
Just apply that to $x$, which has a power of $1$. Remember that any number to the zeroth power is equal to $1$.
Hope that helps. 
A: You ask why the derivative of the function given by $f(x)$ is $f'(x) = 1$. The proper answer is found in the definition. Given a function $f$, the derivative is defined as
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}.
$$
So if $f(x) = x$, then
$$\begin{align}
f'(x) &= \lim_{h\to 0}\frac{\color{blue}{f(x+h)} - \color{red}{f(x)}}{h} \\ &= \lim_{h\to 0}\frac{\color{blue}{x + h} - \color{red}{x}}{h} \\ &= \lim_{h\to 0}\frac{\color{blue}{h}}{h} \\ &= \lim_{h\to 0} 1 = 1.
\end{align}
$$
Now this tells you exactly why the derivative of the function given by $f(x) = x$ is the constant function $1$.
If you want a way to think about it, then remember that the derivative of a function at a number $x$ is the slope of the tangent line to the graph at the point corresponding to $x$. And if you sketch the graph of the function given by $f(x) = x$, then you get a straight line with slope $1$. So at any point, then slope of the tangent line is, in particular, equal to $1$.  
A: You can use the power rule for finding the derivative that,$$\frac{dx^n}{dx}= n x^{n-1}.$$
We know that $x$ is the same thing as $x^1$, and use the above power rule, you get $\frac{dx}{dx}=1 x^0$ and we also know that $x^0$ is $1$, so $\frac{dx}{dx}=1 * 1=1$.
The integral is $$\int x^n dx= \frac{x^{n+1}}{n+1}$$
So the derivative of $x$ is one and the integral of $dx$ is $x$.  If you want to go little different, you can use this formula for finding the derivative, $$\frac{d f(x)}{dx}=
\lim_{h\rightarrow0}⁡ \frac{f(x+h)-f(x)}h$$ you can substitute $x$ in $f(x)$ and still you will get one.
