Meaning of a notation regarding Gradient and the line element In the book 'Spacetime and Geometry' by Sean Caroll, in chapter $1$, he writes that the notation for the gradient is d$\phi$ for some scalar function $\phi$. Then in section 2.4, he uses the expression $$ds^2=g_{\mu\nu}\text{d}x^\mu \text{d}x^\nu$$ and goes on to explain that here $\text{d}x$ here is not a differential element but the gradient(please note the different $d$ and $\text{d}$). My question is if that is the gradient in the equation of line element, then how does the line element equation justify as being correct?
 A: I don't know how familiar you are with differential geometry, so I'll try to give a self contained answer. First, as you might know, the metric tensor $g$ is a bilinear map taking two (tangent) vectors, e.g. velocity vectors. This object can be defined without choosing coordinates of our space(time). Now take coordinate functions $x^0,...,x^d$, then we can introduce a basis $e_\mu$ in the spaces of tangent vectors in a natural way (this basis is often denoted in Terms of differentials $\partial_\mu$). Thus, a velocity vector $v$ for example can be represented as $v=v^\mu e_\mu$. 
However, to get to your question, the gradients of the coordinate functions d$x^\mu$ give you the coefficients of a Vector with respect to the chosen coordinates, i.e. d$x^\mu(v)=v^\mu$. Note that here the gradient d$f$ is not a vector(field), but a linear function taking tangent vectors. A map taking tangent vectors can now be represented in terms of this gradients (i.e. they define a basis). The metric can be expressed as $$g=g_{\mu\nu}\text{d}x^\mu\text{d}x^\nu.$$ So pluging in two tangent vectors $v,w$ gives $$g(v,w)=g_{\mu\nu}\text{d}x^\mu(v)\text{d}x^\nu(w)=g_{\mu\nu}v^\mu w^\nu.$$ Now, the line element $ds^2$ is defined to be the map $$v\mapsto g(v,v),$$ so in coordinates we get essentially the same representation as for the metric tensor, just that we plug in the same vector in both slots. So as a last note, the term "$ds^2$" should be taken as a single term, as you take $\frac{df}{dx}$ as a single term and not as a fraction.
I hope this could answer your question! However, note that this was sloppy in rigour as well as in notation. If you want a nice introduction to this stuff (essentially differential geometry), a nice introduction can be found in M. Nakahara's "Differential geometry, topology and physics" as well as in C. Isham's "Modern differential geometry for physicists".
Cheers!
