3 Contents
2 4 5 Lecture topics 09 - 28 - 98 Whats it mean when the derivative does not exist? Part 1 Whats it mean when the derivative does not exist? Part 2 Whats it mean when the derivative does not exist? Part 3
Lecture topics 09 - 28 - 98
2) Used chain rule to dierentiate (x2 + 1)3 , sin x, sin (cos(x)) Stated chain rule for sin ( ). 3) Used chain rule to dierentiate y = |x2 1|, then to locate on the graph where y = 0, y dne. 4) Talked about meaning of chain rule for [sin(2x)] versus [sin(x)] .
1) Stated chain rule for [f (g (x))] , [f (
)] and
dy dx .
Whats it mean when the derivative does not exist? Part 1
In class I showed that for y = |x2 1| then you can use the chain rule to get x2 1 (2x) y= |x2 1| Then you check that y dne when x = 1. If you look at the graph, you this is two places on the curve where theres pointy places. And that cuz theres pointy places, the function cant have a tangent there. Check the pic:
Whats it mean when the derivative does not exist? Part 2
x This isnt really all there is because the piece of the derivative |x2 1| 1 actually has jumps at x = 1. The point I want to make is the that limits at the jump mean something. Check it out:
2
x1+
lim
x2 1 |x2 1| x2 1 |x2 1|
(2x) =
x1+
lim
x2 1 x2 1 (x2 1) x2 1
(2x) =
x1+
lim (2x) = 2
x1
lim
(2x) = lim
x1
(2x) = lim (2x) = 2
x1
The derivative means something geometrically: its the slope of a tangent line. The function y = |x2 1| has two tangent lines at x = +1; the one at the left has slope slope = 2; the one on the right has slope slope = +2. Check the right pic below.
right tangent; slope = +2
left tangent; slope = -2
Whats it mean when the derivative does not exist? Part 3
Ive made a point about how, when a derivative does not exist, that can tell you something about the ...