Derivative of function f(x)=x|x|

• Aug 9th 2010, 09:12 PM
alexprem
Solved
For the function f(x)=x|x|, show that f'(0) exists. wht is the value?
• Aug 9th 2010, 09:25 PM
pickslides
$f(x)= |x| = x, x\geq 0 \implies f'(x)=1$

$f(x)= |x| = -x, x< 0 \implies f'(x)=-1$

Maybe use the product rule from here?
• Aug 9th 2010, 09:58 PM
Prove It
$|x|$ is not differentiable at $x = 0$, so what makes you think that $x|x|$ is differentiable there?
• Aug 9th 2010, 09:58 PM
Also sprach Zarathustra
x|x| = sgn(x) * x^2
• Aug 10th 2010, 03:06 AM
HallsofIvy
$f'(0)= \lim_{h\to 0}\frac{f(h)- f(0)}{h}$

For f(x)= x|x|, this is $f'(0)= \lim_{h\to 0}\frac{h|h|- 0}{h}= \lim_{h\to 0}|h|= 0$.

Yes, Prove It, even though |x| is not differentiable at x= 0, x|x| is!
• Aug 10th 2010, 04:23 AM
eumyang
Posted by pickslides from this thread:
http://www.mathhelpforum.com/math-he...on-153220.html
Quote:

Wow this is really similar to http://www.mathhelpforum.com/math-he...on-153221.html

Are you in the same class or the same person?
Interesting... Here is another instance of alexprem and ilovemymath posting the exact same question.

http://www.mathhelpforum.com/math-he...ve-153114.html

Coincidence? (Wondering)
• Aug 10th 2010, 10:34 AM
ilovemymath
might be someone from my class
because we are taking this over online so many of us will have problem as we are not being taught anything
• Aug 10th 2010, 11:28 AM
skeeter
Quote:

Originally Posted by ilovemymath
might be someone from my class
because we are taking this over online so many of us will have problem as we are not being taught anything

... such is the major pitfall of "online" classes ... we see many who come here with little or no background in the subject matter for the course they are attempting. I seriously question their academic value.