Hello I have these instructions to solve an instantaneous velocity problem and I don't understand what does it mean by "differentiate the expression" in part b.

(a) Use ∆x/∆t. To calculate the particle's coordinate at the beginning of the interval substitute into the equation for and to calculate its coordinate at the end of the interval substitute . , of course, is the difference and is

(b), (c), and (d) Differentiate the expression for the coordinate with respect to time and evaluate the result for t=2s. Do the same for t=3s and for t=2.50s.

Question

Hello I have these instructions to solve an instantaneous velocity problem and I don't understand what does it mean by "differentiate the expression" in part b. 

(a) Use ∆x/∆t. To calculate the particle's coordinate at the beginning of the interval substitute  into the equation for  and to calculate its coordinate at the end of the interval substitute . , of course, is the difference and  is 

(b), (c), and (d) Differentiate the expression for the coordinate with respect to time and evaluate the result for t=2s. Do the same for t=3s and for t=2.50s.

anonymous · Answer

what's the expression?
differentiate means take the derivative.