There are rules which relate the derivative to the various operations done with vectors such as the dot product, the cross product, vector addition, and scalar multiplication.
Theorem 16.0.9 Let a,b ∈ ℝ and suppose f^{′}
 (16.2) 
 (16.3) 
If f,g have values in ℝ^{3}, then
 (16.4) 
The formulas, 16.3, and 16.4 are referred to as the product rule.
Proof: The first formula is left for you to prove. Consider the second, 16.3.

= lim_{h→0} +


= lim_{h→0} 

= lim_{h→0} ∑
_{k=1}^{n}f_{
k} +
∑
_{k=1}^{n}
g_{k} 

= ∑
_{k=1}^{n}f_{
k}
g_{k}^{′} +
∑
_{k=1}^{n}f_{
k}^{′}
g_{
k} =
f^{′}
⋅ g +
f
⋅ g^{′}
. 
Formula 16.4 is left as an exercise which follows from the product rule and the definition of the cross product. ■
Example 16.0.10 Let r
From 16.4 this equals

Example 16.0.11 Let r
This equals
Example 16.0.12 An object has position

kilometers where t is given in hours. Find the velocity of the object in kilometers per hour when t = 1.
Recall the velocity at time t was r^{′}

When t = 1, the velocity is

Obviously, this can be continued. That is, you can consider the possibility of taking the derivative of the derivative and then the derivative of that and so forth. The main thing to consider about this is the notation, and it is exactly like it was in the case of a scalar valued function presented earlier. Thus r^{′′}
When you are given a vector valued function of one variable, sometimes it is possible to give a simple description of the curve which results. Usually it is not possible to do this!
Example 16.0.13 Describe the curve which results from the vector valued function r
The first two components indicate that for r
As an application of the theorems for differentiating curves, here is an interesting application. It is also a situation where the curve can be identified as something familiar.
Example 16.0.14 Sound waves have the angle of incidence equal to the angle of reflection. Suppose you are in a large room and you make a sound. The sound waves spread out and you would expect your sound to be inaudible very far away. But what if the room were shaped so that the sound is reflected off the wall toward a single point, possibly far away from you? Then you might have the interesting phenomenon of someone far away hearing what you said quite clearly. How should the room be designed?
Suppose you are located at the point P_{0} and the point where your sound is to be reflected is P_{1}. Consider a plane which contains the two points and let r

This reduces to
 (16.5) 
Now

and a similar formula holds for P_{1} replaced with P_{0}. This is because

and so using the chain rule and product rule,

showing that