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Derivative of piecewise function(split in 3 regions)

  1. Aug 12, 2013 #1
    I use the general formula: if y=|f(x)| then y'=[itex]\frac{d|f(x)|}{dx}= \frac{f(x)*f'(x)}{|f(x)|}[/itex] 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
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  3. Aug 12, 2013 #2

    Stephen Tashi

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    That general formula isn't correct.
     
  4. Aug 12, 2013 #3
    Is it correct now?I edited it.
     
  5. Aug 12, 2013 #4

    Stephen Tashi

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    Yes and I think you have the correct answer to the problem.
     
  6. Aug 15, 2013 #5

    HallsofIvy

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    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 [itex]-1\le y< 1[/itex], 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 [itex]y\ge 0[/itex], 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 [itex]\alpha m_1[/itex] if y< -1, [itex]\alpha m_1- (m_0- m_1)[/itex] if -1< y< 1, and [itex]\alpha m_1[/itex] 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.
     
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