What Is The Derivative Of A Sawtooth Function?

Question

What would it's graph look like.

Answer 1

It would look like this: _-_-_-_-_-_-_

where the values of the - and _ are the slope of each side of the tooth (- will be positive, _ will be negative)

Answer 2

Sawtooth function must be a function whose graph looks like a sawtooth. Maybe something like:

/\/\/\/\/\/\/\/\/\/\/\/\/\/\

Now to find the derivative, you need to think in termes of smooth points and cusps (vertices). If you are at a smooth point, i.e. a point on an edge then the function looks like a line around that point and the derivative of the function is the same as the derivative of the line. I am giving you a homework: find the derivative of a line. Now if you are on a sharp point, i.e, a vertex, then there is no derivative. Why that is so is also part of your homework.

Answer 3

Derivative Of Square Wave

Answer 4

A sawtooth wave rises from zero to its maximum value at a constant rate. It then drops abruptly to zero and repeats. Its derivative is a costant except at the discontinuity, where it cannot be defined. Because the derivative cannot be defined at the discontinuities, the function is said not to have a derivative.

Answer 5

Do you mean a triangular (/\/\/\/\/\) or a ramp (/|/|/|/|/|) function? I've heard them both refered to as sawtooth functions.

The derivative of the triangular wave( call it f(t) ) looks like a square wave. f'(t) is constant positive on the upward slopes of f(t) and constant negative on the downward slopes of f(t). It is not a true square wave in that it is undefined on the points where it transitions from positive to negative.

The derivative ( g'(t) ) of the ramp wave ( g(t) ) is a constant: a horizontal line. However, g'(t)it is not defined at the points where g(t) jumps to reset itself.