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Math Help - Differentiating Absolute Values

  1. #1
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    Differentiating Absolute Values

    How do we differentiate f(x,y)=|xy|? I know that the derivative of xy itself is x+y, and the derivative with respect to x and y are respectively f_x=y and f_y=x. But I don't understand how to find the derivatives with the absolute value signs. Can anyone show me?
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  2. #2
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    Consider a separate function

    To make myself clearer I am going to consider the function f(x) = |x|. This function can be rewritten as:

    f(x) = \left\{ \begin{matrix}<br />
x & \mbox{ if } 0 \leq x \\<br />
-x & \mbox{ if } 0 > x \ <br />
\end{matrix} \right.

    So the derivative is:

    f'(x) = \left\{ \begin{matrix}<br />
 1 & \mbox{ if } 0 < x \\<br />
-1 & \mbox{ if } 0 > x \ <br />
\end{matrix} \right.

    It can be shown that in x = 0 the derivative does not exist. Apply this idea to your function.
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  3. #3
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    Quote Originally Posted by Diego View Post
    To make myself clearer I am going to consider the function f(x) = |x|. This function can be rewritten as:

    f(x) = \left\{ \begin{matrix}<br />
x & \mbox{ if } 0 \leq x \\<br />
-x & \mbox{ if } 0 > x \ <br />
\end{matrix} \right.

    So the derivative is:

    f'(x) = \left\{ \begin{matrix}<br />
1 & \mbox{ if } 0 < x \\<br />
-1 & \mbox{ if } 0 > x \ <br />
\end{matrix} \right.

    It can be shown that in x = 0 the derivative does not exist. Apply this idea to your function.
    I kind of knew this definition but I don't see how it helps us to find the derivative... I know that the derivative of |x| is \frac{x}{|x|}. But how do we get this from the definition you've posted?
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  4. #4
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    Apply the definition of derivative to different parts of the function

    Suppose x > 0 then just apply the definition:

     <br />
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}=<br />
\lim_{h \to 0} \frac{x+h - x}{h} =<br />
\lim_{h \to 0} \frac{h}{h}=<br />
\lim_{h \to 0} 1 =<br />
1<br />

    Suppose x < 0 then:

     <br />
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}=<br />
\lim_{h \to 0} \frac{-x-h + x}{h} =<br />
\lim_{h \to 0} \frac{-h}{h}=<br />
\lim_{h \to 0} -1 =<br />
-1<br />

    Now if x = 0

     <br />
\lim_{h \to 0^{+}} \frac{f(x+h) - f(x)}{h}=<br />
\lim_{h \to 0^{+}} \frac{h}{h}=<br />
\lim_{h \to 0^{+}} 1 =<br />
1<br />

    and:

     <br />
\lim_{h \to 0^{-}} \frac{f(x+h) - f(x)}{h}=<br />
\lim_{h \to 0^{-}} \frac{-h}{h}=<br />
\lim_{h \to 0^{-}} -1 =<br />
-1<br />

    thus the derivative on zero does not exist.

    f'(x) = \frac{x}{|x|}, provided x is not zero, is just a compact way of writing the derivative, but is still the same function as:

    f'(x) = \left\{ \begin{matrix}<br />
 1 & \mbox{ if } 0 < x \\<br />
-1 & \mbox{ if } 0 > x \ <br />
\end{matrix} \right.

    Maybe what you wanted to know is if you can obtain the derivative in the form x/|x| by consdering any x, I do not know if this is possible, but in any case I don't think it is that important, because you still obtain the same result. I hope this was the answer you were looking for, if not do tell me.
    Last edited by Diego; April 17th 2010 at 07:20 AM.
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  5. #5
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    Well, you could apply the chain rule.

    Let f=|xy| and u=xy, so f = |u|

    \frac{\partial f}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x} = \frac{u}{|u|} \cdot y = \frac{xy}{|xy|} \cdot y

    = \frac{xy^2}{|xy|} = \frac{x|y|}{|x|}

    xy \neq 0
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  6. #6
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    First let me point out that the statement of the "one thing you know" was wrong. The derivative of xy is NOT xy. If you are differentiating with respect to x and y is a function of x, it is x+ xy'. If you are differentiating with respect to y and x is a function of y, it is y+ yx'. If x and y are both functions of some parameter and you are differentiating with respect to that parameter it is x'y+ xy'. If x and y are independent variable, the partial derivatives are xy_y= x and xy_x= y.

    For |xy|, assuming x and y are independent variables,
    In the first quadrant, x> 0, y> 0 so |xy|= xy and the partial derivatives are |xy|_x= (xy)_x= y and |xy|_y= (xy)_y= x

    In the second quadrant, x< 0 and y> 0 so |xy|= -xy and the partial derivatives are |xy|_x= (-xy)_x= -y and |xy|_y= (xy)_y= -x.

    In the third quadrant, x< 0 and y< 0 so xy> 0, |xy|= xy and the partial derivatives are |xy|_x= (xy)_x= y and |xy|_y= (xy)_y= x.

    In the fourth quadrant, x> 0 and y< 0 so |xy|= -xy and the partial derivatives are |xy|_x= (-xy)_x= -y and |xy|_y= (xy)_y= -x.

    Of course, |xy| is not differentiable if either x or y is 0- that is, on the axes.
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