# Math Help - Differentiating Absolute Values

1. ## 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?

2. ## Consider a separate function

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

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

So the derivative is:

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

It can be shown that in x = 0 the derivative does not exist. Apply this idea to your function.

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

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

So the derivative is:

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

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?

4. ## Apply the definition of derivative to different parts of the function

Suppose x > 0 then just apply the definition:

$
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}=
\lim_{h \to 0} \frac{x+h - x}{h} =
\lim_{h \to 0} \frac{h}{h}=
\lim_{h \to 0} 1 =
1
$

Suppose x < 0 then:

$
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}=
\lim_{h \to 0} \frac{-x-h + x}{h} =
\lim_{h \to 0} \frac{-h}{h}=
\lim_{h \to 0} -1 =
-1
$

Now if x = 0

$
\lim_{h \to 0^{+}} \frac{f(x+h) - f(x)}{h}=
\lim_{h \to 0^{+}} \frac{h}{h}=
\lim_{h \to 0^{+}} 1 =
1
$

and:

$
\lim_{h \to 0^{-}} \frac{f(x+h) - f(x)}{h}=
\lim_{h \to 0^{-}} \frac{-h}{h}=
\lim_{h \to 0^{-}} -1 =
-1
$

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:

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

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.

5. Well, you could apply the chain rule.

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

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

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

$xy \neq 0$

6. 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.