Differential of $x \sin x$ Finding the differential of $x \sin(x)$
\begin{align}
d(x \sin x) &= (dx) sinx + x (d(\sin x)) &&\text{using the product rule}\\
&= \sin x\,dx + x \cos x\,dx
\end{align}
My question is why is the differential of $\sin x$ equal to $(\cos x) (dx)$ and not just $\cos x$ (which is the derivative of $\sin x$.)
A similar question is, why is the differential of $x^2$ equal to $2x(dx)$? Why is there an additional $dx$ in both cases? 

You have a function $f(x)$ and are looking for the differential $df(x)$. In the case $f(x)=x\sin(x)$:
$$\begin{align}
f'(x) &= \frac{df(x)}{dx}\\
df(x) &= f'(x)dx\\
df(x) &= \left( \frac{df(x)}{dx} \right) dx \\
d\left( x \sin (x) \right) &=\left( x \cos(x) + \sin(x) \right)dx
\end{align}
$$
Similarly, for $f(x)=x^2$:
$$\begin{align}
f'(x) &= \frac{df(x)}{dx}\\
df(x) &= f'(x)dx\\
df(x) &= \left( \frac{df(x)}{dx} \right) dx \\
d(x^2) &= (2x)dx
\end{align}
$$
