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? 

in short: $f'(x) = \frac{d f(x)}{dx}$. In your case $f(x) = x \sin x$.
