1. ## Modulus Functions

Hello again! I was hoping someone can help me with dervatives of mod functions, thanks!

Considering suitable domains find the rule for the derivative function:
y = mod(sinx), y = mod(sin2x), y = mod((e^(-0.1x))sinx) and y = mod((e^(-0.2x)sin(3x)).

Consider y = 5(mod(e^(-ax)sin(bx)). What is the effect of changing the value of a and b?

2. Originally Posted by classicstrings
Hello again! I was hoping someone can help me with dervatives of mod functions, thanks!

Considering suitable domains find the rule for the derivative function:
y = mod(sinx), y = mod(sin2x), y = mod((e^(-0.1x))sinx) and y = mod((e^(-0.2x)sin(3x)).

Consider y = 5(mod(e^(-ax)sin(bx)). What is the effect of changing the value of a and b?
y= |sin(x)| = sgn(sin(x)) sin(x)

Where sgn(u)=1 if u>=0 and sgn(u)=-1 if u<0

so between the zeros of sin(x) sgn(sin(x)) is a constant so:

dy/dx = sgn(sin(x)) d/dx sin(x) = sgn(sin(x)) cos(x).

Now we need to check if the derivative exists when sin(x)=0, but we
know that |sin(x)| has corners at these points so the derivative does
not exist;

so:

d/dx[|sin(x)| = sgn(sin(x)) cos(x) for x in R-{k*pi, k in Z}

This may now be simplified by observing that when u!=0 we may
write:

sgn(u) = u/|u|,

so we may write:

d/dx[|sin(x)| = [sin(x)/|sin(x)|] cos(x) for x in R-{k*pi, k in Z}

RonL

3. Hi Ron!

I have a few questions...

Also - you wront mod(sinx) as being equal to (mod(sinx))*sinx, how did you get that?

I have plotted the derivative of y = sgn(x), and I see that it could be said to be a composite function with cos(x) and -cos(x) being defined for particular values of x. However I am having trouble finding those values where they are defined.

I actually have never seen your method, but I think I need to write my derivatives as a composite of two functions where there is no mod at all. Is your method just a simplified version of a possible composite function?

4. Originally Posted by classicstrings
Hi Ron!

I have a few questions...

Also - you wront mod(sinx) as being equal to (mod(sinx))*sinx, how did you get that?
That is not what I did as far as I recall, if I did it is a mistake. What I did
do was define a function sgn(u) which is equal to 1 if u>=0, and -1 if u<0.
Essentialy this is the sign if u.

Then |u| = sgn(u) u, since if u<0 sgn(u)=-1.
(also sgn(u)|u| = u as 1/sgn(u)= sgn(u)).

(here sgn is an abreviation for signum or sign)

RonL

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### domain of mod of log(sinX)

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