Unit 1. Differentiation - aprende.org
ocw.aprende.org › MIT18_01SC_pset1solproblem using the function s(t) = 16t2, representing the distance down measured from the top. Then all the speeds are positive instead of negative.) b) Solve h(t) = 0 (or s(t) = 400) to find landing time t = 5. Hence the average speed for the last two seconds is h(5) − 2h(3) = 0 − (400 − 16 · 3 ) = −128ft/sec 2 2 3
Chapter 5 Techniques of Differentiation
www.math.smith.edu › ~callahan › cic278 CHAPTER 5. TECHNIQUES OF DIFFERENTIATION this general formula agrees with the specific value f′(2) = 12 we have already obtained. Notice the difference between the statements f′(x) ≈ ∆y/∆x and f′(x) = 3x2. For a particular value of ∆x, the corresponding value of ∆y/∆x is an approx-imation of f′(x). We can obtain ...