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

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- Feb 19th 2009, 05:14 PMcalc_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 PMOnMyWayToBeAMathProffesor
I think the formula is $\displaystyle \sqrt{u^2}$ with $\displaystyle u$ being $\displaystyle (x-1)$ but I am not sure how to apply that. Maybe someone else could throw in their input?

- Feb 19th 2009, 05:55 PMskeeter
let u be a function of x ...

$\displaystyle y = |u|$

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

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

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

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

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

=

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

correct? - Feb 19th 2009, 06:10 PMskeeter
no ... you have to use the product rule

- Feb 19th 2009, 06:20 PMOnMyWayToBeAMathProffesor
I don't know why i keep making these simple mistakes. so if we want the derivative it would be $\displaystyle 4x \cdot \frac{x-1}{|x-1|}+2x^2 ?$

I am getting stuck here. - Feb 19th 2009, 06:40 PMskeeter
$\displaystyle y = 2x^2 |x-1|$

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