Understanding directional derivative and the gradient I'm having trouble understanding the proof of directional derivative and the gradient. Could someone give me a easy-to-read proof of the directional derivative and explain why does the gradient point to the direction of maximum increase?
Thank you very much for any help! =)
 A: As for why the gradient points in the direction of maximum increase, let's say we don't know this and we want to find a unit vector $\vec{u}$ such that the directional derivative of some function $f$ is the greatest in the direction of $\vec{u}$. Then if $\theta$ is the angle between $\nabla f$ and $\vec{u}$ we have $\nabla f_{\vec{u}}=\nabla f \cdot \vec{u}=|\nabla f||\vec{u}|\cos(\theta)=|\nabla f|\cos(\theta)$ since $\vec{u}$ is a unit vector. This quantity is then maximized when $\cos(\theta)=1$, i.e., when $\theta=0$, thus $\vec{u}$ points in the direction of the gradient. 
A: As for the gradient pointing in the direction of maximum increase, recall that the directional derivative is given by the dot product
$$\nabla f(x)\cdot\textbf{u},$$
where $$\nabla f(x)$$ is the gradient at the point $\textbf{x}$ and $\textbf{u}$ is the unit vector in the direction we are considering. Recall also that this directional derivative is the rate of increase/decrease of the function in the direction of $\textbf{u}$ at the point $\textbf{x}$. The dot product has two equivalent definitions:
$$\textbf{u}\cdot\textbf{v}=u_1v_1+u_2v_2+...+u_nv_n$$
or
$$\textbf{u}\cdot\textbf{v}=||\textbf{u}||||\textbf{v}||cos(\theta),$$ where $\theta$ is the angle between the vectors.
Using this second definition, the fact that $\textbf{u}$ is a unit vector, and knowledge of trigonometry, we see that the directional derivative $D\textbf{u}$ at $\textbf{x}$ is bounded:
$$-||\nabla f(x)||=cos(\pi)||\nabla f(x)||\leq cos(\theta)||\nabla f(x)||$$
$$\leq cos(0)||\nabla f(x)||=||\nabla f(x)||$$
Since $0\leq||\nabla f(x)||$, we see that the maximum rate of change must occur when $\theta=0$, that is, in the direction of the gradient.
As for the directional derivative consider the direction of the vector $(2,1)$. We want to know how the value of the function changes as we move in this direction at a point $\textbf{x}$. Well, for infinitesimally small units, we are moving $2$ units in the $x$ direction and $1$ unit in the $y$ direction so the change in the function is $2$ times the change in $f$ that we get as we move $1$ unit in the $x$ direction plus $1$ times the change in $f$ that we get as we move $1$ unit in the $y$ direction. Finally, we divide by the norm of $(2,1)$ so that we have the change for $1$ unit in this direction.
A: Although the previous 2 answers are the "real" answers to your inquiry I feel that perhaps you were asking for an intuitive motivation for why grad f represents the maximum increase. The reason lies in the formal reasoning above... Think about what theta being zero really says... that grad f points in the direction orthogonal to (perpendicular) to the contour lines.. higher regions will have contour lines that get closer together and more dense and if grad f if orthogonal to these lines it must be pointing straight up. 
