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? 

$x\sin(x)$ is the product of two functions. It's easier to see that there are two functions here if you rewrite components of the problem as $f(x)=x$ and $g(x)=\sin(x)$. If the two functions $f(x)$ and $g(x)$ are differentiable (the derivative exists), then their product is also differentiable.  
Now apply the product rule: ${(fg)}' = {f}'g+f{g}'$ to get $x\cos(x)+\sin(x)$.
