Derivative of a Continuous Piecewise Function The Question:
Suppose $f: \mathbb{R} \to \mathbb{R}$ and
$$ f(x) =
\begin{cases}
-x^3,  & \text{if $x \le 0$;} \\
x^2, & \text{if $0 < x \le 1$;} \\
\sqrt{x}, & \text{if $1 < x$.}
\end{cases} $$
Does $f$ have a derivative at $x = 1$?
Where I Am:
Well, clearly, $f$ is continuous at $1$, so that satisfies that necessary condition. Furthermore, the LHS limit and RHS limit as $x$ approaches $1$ are both $1$, so that's fairly strong evidence that $f'(1)$ exists and is equal to $1$. Yet, I'm still not convinced, since there's a "sharp point" (edit: it's not "smooth" -- it took me a minute to remember the terminology) at $f(1)$ on its plot (similiar to the absolute value function at at $0$). Is there some sufficient condition for differentiability that I'm overlooking? (It's been a while since calculus.)
 A: $$\lim_{x \to 1^-}\frac{f(x)-f(1)}{x-1}=\lim_{x \to 1^-}\frac{x^2-1}{x-1}=2$$
$$\lim_{x \to 1^+}\frac{f(x)-f(1)}{x-1}=\lim_{x \to 1^+}\frac{\sqrt{x}-1}{x-1}=\frac{1}{2}$$
So from these you can conclude that $f$ is not differentiable at $x=1$
A: Two conditions:
1) $f(x)$ is continuous at $x= a$.  Which is to say that $\lim\limits_{x\to a^-}f(x) = \lim\limits_{x\to a^-} f(x)= f(a)$.
This is a necessary but not sufficient condition which doesn't capture any of the essence of the derivative itself.
2) $\lim\limits{h\to 0^+}\frac{f(x+h) - f(h)}{h}$ (sometimes could the right hand derivative) and $\lim\limits{h\to 0^-}\frac{f(x+h) - f(h)}{h}=\lim\limits{h\to 0^+}\frac{f(x-h) - f(h)}{-h}$ (sometimes could the leftt hand derivative) both exist and are equal.
This is a also a necessary but not sufficient continue; this one does capture the essential definition of the derivative-- except for the fact that the point is question exists and behaves.
...
If $g(x) = x^2$ and $h(x) = \sqrt{x}$ so so $f(x) = g(x); 0<x\le 1$ and $f(x) = h(x)=\sqrt{x}; x > 1$.  $g,h$ are both continuous and differentiable at all positive points.
Condition 1: says for $f(x)$ to be differentiable at $x = 1$ that $g(1) = h(1) = f(1)$.  They do.
Condition 2: says for $f(x)$ to be diffentiable at $x=1$ that $g'(1)=h'(1)$.  They do not. $g'(x) = 2x$ and $h'(x)= \frac 1{2\sqrt{x}}$.  $g'(1)=2 \ne \frac 12 = h'(1)$.  
So $f(x)$ is not differentiable at $x=1$ and it does indeed have a "sharp point" at $(0,1)$ where the curve has a slope of $2$ on one side and then "bends" "sharply" to a curve of $\frac 12$ without "going through" any intermediate slope vales.
