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? 

Depending on how familiar you are with the chain rule and derivative of logarithmic functions, you could differentiate $x \sin{x}$ using this alternate method that doesn't directly use the product rule,
$$\begin{align} y &= x \sin{x} \\
\ln(y) &= \ln( x \sin{x}) \\
\ln(y) &= \ln(x) + \ln(\sin{x}) \\
\frac{1}{y} \frac{dy}{dx} &= \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \\
\frac{dy}{dx} &=  \left( \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \right) \cdot y \\
\frac{dy}{dx} &=  \left( \frac{1}{x} + \frac{\cos{x}}{\sin{x}} \right) \cdot x \sin{x}\\
\frac{dy}{dx} &= \sin{x} + x \cos{x}. \\ \end{align}$$
