# [SOLVED] Derivative of function involving absolute values

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• Feb 19th 2009, 05:14 PM
calc_student09
[SOLVED] Derivative of function involving absolute values
Hi, how would you find the derivative of 2x^2|x-1|
This is calc ab but we haven't learned the derivatives of absolute values
• Feb 19th 2009, 05:26 PM
OnMyWayToBeAMathProffesor
I think the formula is $\sqrt{u^2}$ with $u$ being $(x-1)$ but I am not sure how to apply that. Maybe someone else could throw in their input?
• Feb 19th 2009, 05:55 PM
skeeter
let u be a function of x ...

$y = |u|$

$y = \sqrt{u^2} = (u^2)^{\frac{1}{2}}$

$\frac{dy}{dx} = \frac{1}{2} (u^2)^{-\frac{1}{2}} \cdot 2u \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{u}{\sqrt{u^2}} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{u}{|u|} \cdot \frac{du}{dx}$
• Feb 19th 2009, 06:02 PM
OnMyWayToBeAMathProffesor
so if the original equation was $2x^2(|x-1|)$

the derivative would be $2x^2 \cdot \frac{x-1}{|x-1|} \cdot 1$
=

$\frac{2x^3-2x^2}{|x-1|}$

correct?
• Feb 19th 2009, 06:10 PM
skeeter
no ... you have to use the product rule
• Feb 19th 2009, 06:20 PM
OnMyWayToBeAMathProffesor
I don't know why i keep making these simple mistakes. so if we want the derivative it would be $4x \cdot \frac{x-1}{|x-1|}+2x^2 ?$

I am getting stuck here.
• Feb 19th 2009, 06:40 PM
skeeter
$y = 2x^2 |x-1|$

$\frac{dy}{dx} = 2x^2 \cdot \frac{x-1}{|x-1|} + 4x|x-1|$
• Feb 19th 2009, 07:11 PM
OnMyWayToBeAMathProffesor
Thank you very much!