This demo illustrates the Chain Rule for three variables: We have a smooth green curve C given by (x(t),y(t),z(t)) in the three-dimensional domain, and you can use the the tapedeck variable t_0 to specify a point P = (x(t₀),y(t₀),z(t₀)) on the curve. The 3D Graph-Window then shows the tangent vector at P (brown) and the x,y and z-components of the tangent vector. The x-component (x'(t),0,0) is shown in red, the y-component (0,y'(t),0) is shown in blue and the z-component (0,0,z'(t)) is shown in yellow.

The 3D Graph-Window then shows a small part of the curve f(C) = (x(t₀),y(t₀),z(t₀)). (Strictly speaking, the curve would have to lie in four-space.) The point f(P) is shown in grey. The tangent vector at the point f(P) is then shown in brown, with the contributions of the x-,y- and z-components again shown in red, blue and yellow. The three Projection into..-Windows recall how the Chain Rule works in the xy-,xz- and yz-slices, respectively:

For example, take a look at the window titled Projection into the xz-slice. Ignore the z-component for a moment, and recall how we demonstrated the Chain Rule in two dimensions. Then you can see that the contributions from x- and y to the velocity vector at f(P) are given by f_x(P)x'(t₀) and f_y(P)y'(t₀), respectively. Likewise, the window titled Projection into the xz-slice, shows that the contributions from x- and z to the velocity vector at f(P) are given by f_x(P)x'(t₀) and f_z(P)z'(t₀).

Now a simple argument will show that to get the actual velocity vector at f(P), you just have to add the three distinct contribution from the coordinate components of the curve C, and this shows that the velocity vector at f(P) is given by (x'(t₀),y'(t₀),z'(t₀),f_x(P)x'(t₀) + f_y(P)y'(t₀) + f_z(P)z'(t₀)).