This Demonstration visually explains the theorem stating that the directional derivative

of the function

at the point

,

) in the direction of the unit vector

is equal to the dot product

of the gradient of

with

. If we denote the partial derivatives of

at this point by

and

and the components of the unit vector

by

and

, we can state the theorem as follows:

In this Demonstration there are controls for

, the angle that determines the direction vector

, and for the values of the partial derivatives

and

. The partial derivative values determine the tilt of the tangent plane to

at the point

,

); this is the plane shown in the graphic. When you view the "directional derivative triangle", observe that its horizontal leg has length 1 (since

is a unit vector), and so the signed length of its vertical leg repesents the value of the directional derivative

. When you view the "partial derivative triangles", this signed vertical distance is decomposed as the sum

. The first summand is represented by the vertical leg of the blue triangle; the second is represented by the vertical leg of the green triangle. The visual representation is most clear when the two summands have the same sign.