Directional Derivatives and the Gradient

Given a surface of the form z = f(x,y) and a two dimensional unit vector u, the directional derivative is the rate of change of f in the direction of u. It is also the slope of the tangent line of the curve formed by the intersection of the surface with the plane through u that is perpendicular to the xy plane.
We use z = 2-x²-y², the point (0.5,0.5), and the vector (1,1) to illustrate this.
View Surface.

We use the point (0.5,0.5). Note that f(0.5,0.5) = 2-(0.5)²-(0.5)² = 1.5.
View Point.

The gradient is the direction in which the slope of a line tangent to the surface will have the largest absolute value. The gradient is defined to be f_xi + f_yj. For our example, the gradient is -2xi - 2yj. Evaluating this at (x,y)=(0.5,0.5), we get -i - j. Thus a tangent line on the surface z = f(x,y) = 2-x²-y² in the direction -i - j will have the largest possible slope in terms of absolute value.
View Gradient.

We want to know the rate of change of f in the direction (1,1). To make this into a unit vector, we divide by the length of (1,1), namely sqrt(1²+1²) = sqrt(2). Thus u = (1/sqrt(2),1/sqrt(2)).
View Vector.

We next consider the plane which is perpendicular to the xy plane, is parallel to u, and passes through (0.5,0.5).
View Plane.

This plane intersects the surface f in a curve. The parametrization of this curve is (t,t,2-t²-t²) since x = y where the plane intersects the xy plane.
View Curve.

The tangent to this curve is the derivative of the curve, namely (1,1,-2t). Let t = 0.5 to find the tangent at the point (0.5,0.5,1.5). We get that the tangent vector is (1,1,-1).
View Tangent.

The directional derivative gives the slope of the tangent line. The directional derivative is the dot product of the gradient of f and u. The gradient is f_xi + f_yj, which in our case is -2xi - 2yj. Substitute x = 0.5 and y = 0.5 to get -i -j. Then the dot product with u is -1*1/sqrt(2) - 1*1/sqrt(2) = -sqrt(2). Thus the slope is about -1.4.