What is the derivative of $x\sin x$? Ok so I know the answer of $\frac{d}{dx}x\sin(x) = \sin(x)+ x\cos(x)$...but how exactly do you get there? I know $\frac{d}{dx} \sin{x} = \cos{x}$. But where does the additional $\sin(x)$ (in the answer) come in? 

A bit of intuition about the product rule:
Suppose that you have a rectangle whose height at time $t$ is $h(t)$ and whose width at time $t$ is $w(t)$. Then the area at time $t$ is $A(t)=h(t)w(t)$.  Now, as the time changes, how does the area change?

(Please, forgive my use of paint here.)
Say the white rectangle was from time $t$, and the larger rectangle at time $t+\Delta t$. We gain three new regions of area: the green one, the blue one, and the gray one.
The green area is $\Delta h\cdot w(t)$, where $\Delta h$ is the change in height from time $t$ to time $t+\Delta t$; the blue area is, similarly, $\Delta w\cdot h(t)$, and the gray area is $\Delta h\cdot\Delta w$. So, we have
$$
\Delta A=\Delta h\cdot w(t)+\Delta w\cdot h(t)+\Delta h\cdot\Delta w.
$$
Now, when $\Delta t$ is really small, we expect $\Delta h$ and $\Delta w$ to be really small as well; so, their product is tiny. Hence
$$
\Delta A\approx \Delta h\cdot w(t)+\Delta w\cdot h(t),
$$
or
$$
\frac{\Delta A}{\Delta t}\approx\frac{\Delta h}{\Delta t}\cdot w(t)+\frac{\Delta w}{\Delta t}\cdot h(t).
$$
Does this look at all like the product rule?  Letting $\Delta t\rightarrow0$, this approximation (properly formalized, of course) leads us to the formula
$$
\frac{d}{dt}\left[w(t)\cdot h(t)\right]=\frac{dA}{dt}=\frac{dh}{dt}\cdot w(t)+\frac{dw}{dt}\cdot h(t)
$$
