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what is the di/dx of y = cot^-1 (tan2x)? a. -2 b. 2 c. tan2x d. -cot 2x
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what is a di?
i think dy/dx
ops dy/dx sorry
:) i thought so to but can never be too sure
w8 i check the other question i think i put di/dx LOL i fix it
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y = cot^-1 (tan2x) re write it as cot(y) = tan(2x) and derive
-csc^2(y) y' = 2 sec^2(2x) y' = -2 sec^2(2x)/csc^2(y)
iron out the y spot and simplify
i get -2 here
then with any luck, its -2 :)
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thanks god bless
its best to iron out the y spot tho to be sure ... and if you have already done that then I trust your judgement
the wolf agrees ;)
i fix the question here sorry http://openstudy.com/study#/updates/4f5df87fe4b0602be438c4be sorry before
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