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Find f'(x), f(x) = (x^2 - 2x - 6)^3
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Just don't forget your chain rule and you should be fine.
Did you try it yet?
ok, i'll try it, just a sec
Any luck?
i got (2x - 2) and 3u^2, and now?
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OH, WAIT! i'm a idiot. just a sec.
\[3(x^2-2x-6)^2(2x-2)\] Looks right to me (assuming you have u = x^2-2x-6 )
ahh, ok, but now i'm stuck.
what's the next step?
that's it
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there is no way to simplify?
No good simplification. The quadratic is not nicely factored and it has that nasty square, so I'd just leave it that way.
Ahhhh! THANKS! :D
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