TITLE: Is there a simple way of showing that the directional derivative is the dot product of grad(f) with the directional vector u?
QUESTION [6 upvotes]: I am reading a sort of technical proof but I would like to prove it myself in a cleaner and shorter way.
So, I want to show 
$$D_uf = \nabla f. u$$
Any suggestions are greatly appreciated.  
Thanks,
EDIT:  I think I will try to work backwards, using matrix multiplication, since grad(f) can be viewed as a 1x2 matrix.

REPLY [3 votes]: For given point $p$ and direction $u$, $$f(x_p+tu_x,y_p+tu_y,z_p+tu_z)=\phi(t)$$ is a function of $t$ that indicates how $f$ evolves along the straight line through $p$ in the direction $u$.
Then by the chain rule,
$$\frac{d\phi}{dt}=\frac{\partial f}{\partial x}u_x+\frac{\partial f}{\partial y}u_y+\frac{\partial f}{\partial z}u_z=\nabla f\cdot u.$$