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Mathematics 19 Online
OpenStudy (anonymous):

Find the derivative of____(below) and indicate points at which f'(X) does not exist.

OpenStudy (anonymous):

|dw:1379624252849:dw|

zepdrix (zepdrix):

Umm I have an approach, maybe this is over-complicating it though.. Recall: \(\Large |x|\quad=\quad \sqrt{x^2}\) So we can write our expression as:\[\Large f(x)\quad=\quad \sqrt{\left(\sqrt{x^2}-1\right)^2}\] From there, we can take the derivative and apply the chain rule a couple times.

OpenStudy (anonymous):

oh ok, got it thanks!!

zepdrix (zepdrix):

Oh figured it out already? XD

OpenStudy (anonymous):

yeah,wait a sec. let me double check my answer,

zepdrix (zepdrix):

That approach seems fine. Maybe this makes more sense though. Note that:\[\Large \left(|x|\right)' \quad=\quad \frac{x}{|x|}\](You could use the sqrt thing to show this fairly easily). Which will lead us to this:\[\Large \left(\left||x|-1\right|\right)' \quad=\quad \frac{|x|-1}{\left||x|-1\right|}(|x|-1)'\]

zepdrix (zepdrix):

Bah! So many absolute lines lol.

zepdrix (zepdrix):

\[\Large \left(\left||x|-1\right|\right)' \quad=\quad \frac{|x|-1}{\left||x|-1\right|}\left(\frac{x}{|x|}\right)\]

OpenStudy (anonymous):

I'm confused the correct answer is:|dw:1379626117183:dw| I did wrong.. can you show/explain me how to get this answer?? btw, it's not a homework, i'm just practicing and I found this question somewhere in my book.

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