The differential of |cos(x)|?

Question

I am trying to draw a graph of the differential of the absolute value of cos(x), but I don't know how it looks. Is it the graph of |-sin(x)| i.e. |sin(x)| or something different? And at points x= Pi/2 and 3Pi/2 are there point-gaps in the line where it isn't defined?

Answer 1

You are correct in that it is undefined at these points.

It is the graph of -sinx, but only between -pi/2 and pi/2

It repeats at these intervals, so at -pi/2 for example, the curve will be at a value of 1 and -1.

To help you, draw the |cosx| curve and look at what the gradient would be.

The differential will end up looking like a row of backwards S shapes.

Answer 2

The first thing to do it to graph out (in rough terms) the |cosx|. When you do, you see that the derivative is not going to always be positive, nor is it going to be continuous. From 0 to pi/2, it will be -sinx, but from pi/2 to 3pi/2, it will be sinx, then -sin x again from 3pi/2 to 2pi. You are correct that there are gaps at pi/2 and 3pi/2.

Note that any answer with an absolute value sign in it is clearly wrong. The slope of the |cosx| curve is both positive and negative at different times, so the answer is not always positive or negative.

Answer 3

differential coefficient of
[abs(cosx)]
=abs[-sinx]
=xinx
cosx graph is similar to sinx graph.
When x=0
sinx=0 therefore sinx graph passes through origin
when x=0
cosx=1 In this case the graph has a phase difference of pi/2 or 90Deg

Answer 4

I have plotted the graph in MathCad2006 and it shows the following:

d/dx (|cos x|) = -sin x *(|cos x|)/cos x

i.e.
d/dx (|cos x|) = -sin x for -pi/2 d/dx (|cos x|) = sin x for pi/2 d/dx (|cos x|) = -sin x for 3pi/2 and so on

and since (|cos x|)/cos x is undefined when cos x = 0, then the derivative is undefined at pi/2, 3pi/2 etc.

Fraser.

Answer 5

It will not be a continuous function, there will be a step change from

-sin(x) to +sin(x) or vice versa at points where cos(x)=0

at points where cos(x)>0 it will be -sin(x)
at points where cos(x)<0 it will be +sin(x)

Answer 6

|cos(x)| = - cos x, for, cosx <0, OR, cos x, for cosx >0

so, d/dx ( |cos(x)| ) = sin x, for cosx <0, OR, -sinx for, cosx >0
= -|sin(x)|

plug in x= Pi/2 into this formula to find results, then again or the other value.

Answer 7

Yes cos X becomes -sin X
Try this webpage i think it will help you!

http://www.sosmath.com/calculus/diff/der03/der03.html

Answer 8

Basically it will be -sin(x) where cos(x) is positive, sin(x) where cos(x) is negative, and undefined, as you suggest, where cos(x) = 0.
So I suppose it's -sin(x) * (cos(x)/|cos(x)|).