# taking partial derivatives of a modulus

Could anyone help me with this problem . Evaluate

$\displaystyle\frac{\partial }{\partial x}$ |xy|

$\displaystyle\frac{\partial }{\partial y}$ |xy|

What i did was to take a point (a,b), then i substitute it to the expression above. Hence getting

$\displaystyle\frac{\partial }{\partial x}$ |xb| at x=a

$\displaystyle\frac{\partial }{\partial y}$ |ay| at y=b

However im unsure of how to take partial derivatives invloving a modulus. I believe that we must consider a few cases. If so how to consider those cases?

This is my approach, using definition of partial derivatives:

$\lim_{h\to0} \frac {f(a+h,b)-f(a,b)}{h}$ $= \lim_{h\to0} \frac {|(a+h)b|-|ab|}{h}$ $= \lim_{h\to0} \frac {|(a+h)||b|-|a||b|}{h}$ $= \lim_{h\to0} |b| \frac {|(a+h)|-|a|}{h}$

Now consider the cases. If $a>0$, then $a+h>0$ which gives the limit as $|b|$

If $a<0$, then $a+h<0$ which gives the limit as $-|b|$

• Yup thanks for ur answer. But there is still one more case to consider when a=0 Aug 28, 2014 at 11:23
• In that case, it's $\lim_{h\to0} |b| \frac{|h|}{h}$ and so on.
– Diya
Aug 28, 2014 at 11:28
• The derivative dosent exists at a=0 because the limit does not exists at that point. Aug 28, 2014 at 11:29
• Does this function has a tangent plane at (0,0)?
– user178318
Sep 29, 2016 at 4:53
• I would add, $a+h\overset{(<)}{>}0$ for sufficiently small $h$. Dec 16, 2021 at 17:11

Hint. Can you draw the graph of $f(x)=5|x|$, and then work out the ordinary (not partial) derivative of $f$? That is, state from the graph where the derivative does not exist, and what is its value when it does exist?

If you can do this, then you should be able to find the partial derivative $$\frac{\partial}{\partial x}|xy|\ ,$$ because to do this you just take $y$ to be a constant and then differentiate in the normal manner with respect to $x$.

Another hint. Take particular care with the case when $y=0$.

• that means for the function f(x)=5|x| f'(x)=5 when x>0, f'(x)=-5 when x<0 while f'(x) does not exists when x=0. Am i correct? Aug 28, 2014 at 11:11
• Got it. Now just replace $5$ by $|y|$, and don't forget to look carefully at the case $y=0$. Aug 28, 2014 at 11:22