# Math Help - derivative

1. ## derivative

How do I find the derivative of y = modulus(sin x)? ie. y = absolute(sin(x)). After that I would like to plot both y and the derivative of it. Also, how can I find where y is not differentiable? Can I do it both graphically and algebraically? For the graphic method, can I do it by observation? Does continuity come into play here? I am not sure if contininuity basically means that there is no break in the graph and whether I do a visual inspection to determine continuity and henceforth whether y is differentiable or not.(I am sorry if I am not expressing myself well here)

I know that modulus of sin(x) looks like sin(x) but with all the negative bits of it inverted. I am not really sure about the graph of the derivative though.

How do I do the same things as above for sin of modulus of x? ie. y = sin(absolute(x))

2. Originally Posted by BrainZero
How do I find the derivate of y = modulus(sin x)? ie. y = absolute(sin(x)). After that I would like to plot both y and the derivative of it. Also, how can I find where y is not differentiable? Can I do it both graphically and algebraically? For the graphic method, can I do it by observation? Does continuity come into play here? I am not sure if contininuity basically means that there is no break in the graph and whether I do a visual inspection to determine continuity and henceforth whether y is differentiable or not.(I am sorry if I am not expressing myself well here)

I know that modulus of sin(x) looks like sin(x) but with all the negative bits of it inverted. I am not really sure about the graph of the derivative though.

How do I do the same things as above for sin of modulus of x? ie. y = sin(absolute(x))
note that $y = |\sin x| = \left \{ \begin{array}{lr} \sin x & \mbox{ if } \sin x \ge 0 \\ & \\ - \sin x & \mbox{ if } x < 0 \end{array} \right.$

from the graph, we know the function is not differentiable at a point if there is a cusp, sharp turn, or the function is undefined or discontinuous.

algebraically, we know a function is not differentiable by going to the definition. a function $f(x)$ is differentiable if the limit

$\lim_{h \to 0} \frac {f(x + h) - f(x)}h$

exists. otherwise, it is not differentiable.

for example, $|x|$ is not differentiable at x = 0, since

$\lim_{h \to 0} \frac {|0 + h| - |0|}h = \lim_{h \to 0} \frac {|h|}h = \left \{ \begin{array}{lr} 1 & \mbox{ if } h \ge 0 \\ & \\-1 & \mbox{ if } h < 0 \end{array} \right.$

3. You have to split it into two bits: get the driv of y = sinx and that applies to x > 0, then get the driv of y = -sinx and that applies to x < 0. Investigate what happens at x=0 to see whether it's diffable at this point (I think not, it's got a kink in it).

What you find is y is continuous but y' is not because there's a jump from -1 to +1 and there's no value at x=0.

Same thing applies to sin (modulus x). Do it for +x then do it for -x and you find it comes to the same thing. But then you knew that because everyone knows that sinx is an odd function.

UPDATE: Sorry, I believe that a lot of what's written in here is incorrect. Disregard it.

4. note, there are several...ok, bad word, there are infinitely many points where there are kinks in the graph. sine is periodic, so they occur at regular intervals. they shouldn't give you much trouble to find

5. If I do it piece-wise like you guys have mentioned in your post I get the same derivatives for both y = modulus(sin x) and y = sin(modulus(x)). How is this possible?

I get cos x for x>=0 and -cos x for x<0 for the two functions.

Somebody mentioned using the chain rule to differentiate the functions. I just can't figure for the life of me how to do it?

6. y = modulus (sin x) is a sine wave with *all* the negative bits going positive. Like when you pass AC through a full-wave rectifier.

y = sin (modulus x) is different, it's a conventional sine wave when x is greater than 0, and an upside-down version of this (or mirror-reflected in the y axis, however you want to look at it) when x is less than 0.

Sketch it and see what I mean.