**Problem : **
Find the derivative of *f* (*x*) = 3*x*^{4} -2*x*^{2} +5*x*^{-1} and evaluate it at *x* = 2.

Using the basic calculus rules established in this section, we find that

*f'*(*x*) = 12*x*^{3} -4*x* - 5*x*^{-2} and*f'*(2) = 96 - 8 - 5/4 = 86 + 3/4

**Problem : **
Find the velocity and acceleration functions corresponding to the position
function *x*(*t*) = 3*t*^{2} - 8*t* + 458.

*v*(*t*) = *x'*(*t*) and

*a*(*t*) = *v'*(*t*) = *x''*(*t*), so using our basic calculus rules again we
find that

*v*(*t*) = 6*t* - 8 and *a*(*t*) = 6

Notice that the acceleration in this case is constant, and that its value is
equal to twice the coefficient of

*t*^{2} in

*x*(*t*).

**Problem : **
What happens when a car which is traveling along at constant velocity
screeches to a halt?

The velocity of the car decreases rapidly, corresponding to a large negative
acceleration (or

*deceleration*) of the vehicle (courtesy of good
brakes). While the car was traveling at constant velocity, on the other
hand, the acceleration was zero.