Is the domain of a function necessarily the same as that of its derivative? As in the title, I was wondering if it is necessarily true that the domain of a function is shared by its derivative.
 A: No.
Take for instance $f(x)=\sqrt x$.
The domain of $f$ is $\mathbb R_+$.
But $f'(x)=\frac 1{2\sqrt x}$ which has $\mathbb R_+^*$ for domain.
A: Yes or no, depending on how you view it.  
The usual approach in mathematics is to say that if $E$ is a subset of the real line $\mathbb R$ then the domain of a function $f\colon E\to \mathbb R$ is the set $E$.  Then the derivative of $f$ is only defined if $f$ is differentiable at every point in $E$.  In that case, we can define the derivative $f'\colon E\to\mathbb R$.  Clearly, the domain of $f'$ is $E$ also.  
There are functions $f\colon E\to \mathbb R$ that are not differentiable at every point in $E$ - the function $f(x)=|x|$ is an example.  Technically, such a function has no derivative, so it does not make sense to talk about the domain of its derivative.  
The definition of domain used in schools is different and a bit less precise.  Normally, we have some formula and we say that its domain is the largest set of real numbers such that this formula makes sense, according to various rules (e.g., $1/x$ does not make sense if $x=0$, $\sqrt x$ does not make sense if $x<0$ and so on).  According to this definition of domain, the $f(x)=\sqrt x$ example given by E. Joseph is a good example of a function whose derivative has a smaller domain than that of the original function.  
I don't know if $f(x)=|x|$ counts as an example since its derivative is not given by an explicit formula obtained by applying the rules for differentiation.
A: A more extreme example: the Weierstrass function
$$f(x) = \sum_{n=1}^{\infty}a^n\cos(b^n\pi x)$$
($0<a<1$, $b$ positive odd integer, $ab > 1 + 3\pi/2$)
has
$$\text{dom}(f) = \Bbb R,\qquad\text{dom}(f') = \emptyset.$$
A: No, it is enough to take $f(x)=|x|.$
