Directional derivative vs. function restriction and then derivative Say I have a function of two variables, and a line in the plane, and I'd like to "take the derivative along the line". Is this an indication to use the directional derivative, OR is it expected that I first restrict the function to the line in the plane, so that it's a function of one variable, and then take a derivative in, say, the x variable? 
And whichever one is the right answer, what is the other option doing then?
(this specifically is in relation to a hermite finite element example; the line in question is one side of a triangle, so two points on the line are given (two of the triangle vertices); if the directional derivative is what is in fact meant, then I guess they mean to evaluate it at one of the two points.)
 A: The two methods give the same answers (although there are some subtleties).  For a vector $v=(v_x,v_y)$ and point $x=(x_0,y_0)$, the directional derivative is 
$$
\lim_{t\rightarrow 0}\frac{f(x+tv)-f(x)}{t}.
$$
A standard theorem in multivariate calculus is that the directional derivative is $\nabla f\cdot v=v_x\partial_xf+v_y\partial_yf$.
Now, if you restrict to the line $y=m(x-x_0)+y_0$ (for a non-vertical line), then the derivative of $f(x,y)$ along this line is the derivative of 
$$
f(x,y)=f(x,m(x-x_0)+y_0)
$$
(for now, we'll focus on the derivative with respect to $x$).  The derivative of this (using the chain rule) is 
$$
\frac{d}{dx}f(x,m(x-x_0)+y_0)=\partial_xf+m\partial_yf.
$$
We can now relate the two expressions.  Note that the line $y=m(x-x_0)+y_0$ has direction vector $(1,m)$, so this connects the two methods.
The main caution is that by writing a line as $y=mx+b$, as a function of $x$, this is not a curve of unit speed (one must divide by $\sqrt{1+m^2}$).  Therefore, unless you change your variables, you'll get a scaling error between the directional derivative and the restriction.
