Derivatives and what is a good definition? So I have a question in general about derivatives.  I understand that the formal definition is something like $f$ is differentiable at $x=a$ if the limit exists where that limit is either the limit as $x$ approaches $a$ or $h$ approaches $0$ of the equation etc, etc.  However then we say $f$ is differentiable at $a$.  But when someone asks to find $f'(x)$, are we just taking the derivative like we did way back when in Calculus?  
For example, I am trying to show that sin(1/x) is differentiable for all x, except for when x=0.  I have figured out 1/x is differentiable, but how would I go about finding sin(1/x) is differentiable at x=a, where $a$ is not equal to 0? Should I just use the formal definition? 
 A: For the points where $\sin(\frac{1}{x})$ is differentiable, you don't need to use the formal definition of the derivative. You can instead combine the following theorems about derivatives


*

*$\frac{d}{dx}x^k = kx^{k-1}$ for $k \in \mathbb{Z} \setminus \{0\}$. For $k \leq 0$, the derivate exists everywhere except at $x=0$, for $k > 0$ it exists everywhere.

*$\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$ where $f'(x) = \frac{d}{dx}f(x)$, $g'(x) = \frac{d}{dx}g(x)$, and assuming that $f'$ exists at $g(x)$ and $g'$ at $x$.

*$\frac{d}{dx}\sin(x) = \cos(x)$


Those together yield, as you know from calculus of course, that $$
  \frac{d}{dx}\sin(\tfrac{1}{x}) = -\frac{1}{x^2}\cos(\tfrac{1}{x}) \quad \text{where $x \neq 0$.}
$$
So now you'll just have to show that $\sin(\frac{1}{x})$ does indeed have no derivative at $x=0$.
A: On the whole the formal definitions make rigorous what we first did and also help us to understand the precise conditions when we are safe to use particular theorems.
Normally, shortly after defining the derivative, theorems are proved for suitable differentiable functions such as $$(f+g)'=f'+g'$$ $$(fg)'=f'g+fg'$$ $$(g(f))'=f'\cdot g'(f)$$
So if you have proved the chain rule, for example, you can use it in your example - and either know, or show separately that the sine and reciprocal functions are differentiable (except at zero). At zero you have the challenge of showing the limit doesn't exist.
