Calculus 1 -- Concavity; second derivative test -- More practice

0:00 Introduction
0:39 Problem 1
8:38 Problem 2
19:47 Problem 3
32:32 Problem 4
38:53 Problem 5
48:24 Problem 6
58:33 Problem 7
1:09:55 Problem 8
1:17:20 Problem 9
1:28:27 Problem 10

[jana]

19:56 if anyone is wondering about why ∂/∂x ln|x| = 1/x and we don't have to worry, you can note that (ln|x+h| - ln|x|)/h when x > 0 is just (ln(x+h) - ln(x))/h which tends to 1/x as h -> 0. When x < 0 and h is small, it is (-1)(ln(-x-h)) - ln(-x))/-h and as h -> 0, this tends to (-1)1/(-x) = 1/x. So ∂/∂x ln|x| = 1/x.

[thomaswburkhart]

Coming back and watching Calc1 concepts being explained after it’s been two years since I took that course is therapeutic