Derivative of piecewise function(split in 3 regions)

1. Aug 12, 2013

marellasunny

I use the general formula: if y=|f(x)| then y'=$\frac{d|f(x)|}{dx}= \frac{f(x)*f'(x)}{|f(x)|}$ to calculate the derivative of the piecewise function given below.

Given: Piecewise function
$$h(y)=\alpha(m_1 y+\frac{1}{2}(m_0 - m_1)(|y+1|-|y-1|))$$

My attempt at calculating the derivative:
$$\frac{\mathrm{d} h}{\mathrm{d} y}=\alpha m_1+\alpha 0.5(m_0-m_1)(\frac{y+1}{|y+1|}-\frac{y-1}{|y-1|})$$, is this right?

Last edited: Aug 12, 2013
2. Aug 12, 2013

Stephen Tashi

That general formula isn't correct.

3. Aug 12, 2013

marellasunny

Is it correct now?I edited it.

4. Aug 12, 2013

Stephen Tashi

Yes and I think you have the correct answer to the problem.

5. Aug 15, 2013

HallsofIvy

Staff Emeritus
Another way to look at this is to separate it unto three parts (that is, treat it as being "piecewise" as you say in your title. If y< -1, then both y+ 1 and y- 1 are negative so |y+1|- |y- 1|= -(y+1)-(-(y- 1)= -y- 1+ y- 1= -2. That is a constant so its derivative is 0. If $-1\le y< 1$, then y+1 is positve but y- 1 is still negative. |y+ 1|- |y- 1|= y+ 1- (-(y-1))= y+ 1+ y- 1= 2y. The derivative of that is 2. Finally, if $y\ge 0$, both y+ 1 and y- 1 are positive so |y+ 1|- |y- 1|= y+ 1- y+ 1= 2. Again the derivative of that is 0.

So the derivative of the original function is $\alpha m_1$ if y< -1, $\alpha m_1- (m_0- m_1)$ if -1< y< 1, and $\alpha m_1$ if y> 1. Of course, the function is not differentiable at -1 or 1.

Looking at your proposed derivative, if y< -1, so that both y-1 and y+ 1 are negative, the last part is "-1- (-1)" which is 0 while if y> 1, so that y-1 and y+ 1 are positive, it is "1- 1". If -1< y< 1, then it is "1- (-1)= 2" which cancels the "1/2" so, yes, that is the same thing.