Respuesta :

sqdancefan
Given
  h(t) = 5t² . . . . ft for t in seconds

Find
  h'(5)

Solution
Differentiating, we have
  h'(t) = 10t
Substituting t=5 gives
  h'(5) = 10·5 = 50 . . . . ft/s

5 sec after liftoff, the rocket is climbing 50 ft/s.