[SOLVED] Derivative of function involving absolute values

**calc_student09** · February 19th 2009, 05:14 PM

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

**OnMyWayToBeAMathProffesor** · February 19th 2009, 05:26 PM

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?

**skeeter** · February 19th 2009, 05:55 PM

let u be a function of x ...

$y = |u|$

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

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

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

$\frac{dy}{dx} = \frac{u}{|u|} \cdot \frac{du}{dx}$

**OnMyWayToBeAMathProffesor** · February 19th 2009, 06:02 PM

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?

**skeeter** · February 19th 2009, 06:10 PM

no ... you have to use the product rule

**OnMyWayToBeAMathProffesor** · February 19th 2009, 06:20 PM

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.

**skeeter** · February 19th 2009, 06:40 PM

$y = 2x^2 |x-1|$

$\frac{dy}{dx} = 2x^2 \cdot \frac{x-1}{|x-1|} + 4x|x-1|$

**OnMyWayToBeAMathProffesor** · February 19th 2009, 07:11 PM

Thank you very much!

Thread: [SOLVED] Derivative of function involving absolute values

LinkBack

Thread Tools

Display

[SOLVED] Derivative of function involving absolute values

Similar Math Help Forum Discussions

Derivatives involving absolute values

Find the absolute maximum and absolute minimum values of the function

limit involving absolute values

Integers involving absolute values

Limit involving Absolute Values.

Search Tags