# Thread: Derivative of abs function

1. ## Derivative of abs function

What is the derivative of this function:

$\displaystyle abs(\frac{x}{1+x^2})$

This is the shot I took:

$\displaystyle \frac{x}{1+x^2}$ and $\displaystyle \frac{-x}{1+x^2}$
so $\displaystyle \frac{(x-1)(x-1)}{(x^2+1)^2}$ and $\displaystyle \frac{(x+1)(x+1)}{(x^2+1)^2}$ would be the derivative of those functions respectively

But I really don't understand the fundamentals of this problem - the absolute value is really throwing me and I don't know how to properly deal with it. Thanks so much if you can help!

2. $\displaystyle \left|\frac{x}{x^2+1}\right| = \frac{x}{x^2+1}$ for $\displaystyle x \ge 0$

$\displaystyle \left|\frac{x}{x^2+1}\right| = \frac{-x}{x^2+1}$ for $\displaystyle x < 0$

for $\displaystyle x \ge 0$... $\displaystyle f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$

for $\displaystyle x < 0$ ... $\displaystyle f'(x) = -\frac{1-x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}$

3. Thanks a lot! I guess what I really needed was a little reassurance (and maybe someone to check my algebra!!)