# 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

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.