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OpenStudy (anonymous):

when y = ab Tan^-1 (acotx)/b, then y' is equal to? a. (a^2b^2)/(b^2 sin^2 x + a^2 cos^2 x) b. (-a^2b^2)/(b^2 sin^2 x + a^2 cos^2 x) c. (a^2b^2)/(b^2 cos^2 x + a^2 sin^2 x) d. (-a^2b^2)/(b^2 cos^2 x + a^2 sin^2 x)

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OpenStudy (anonymous):

\[y=ab \tan^{-1} (\frac{a \cot x}{b})\]Is this your function? a and b are constants, I assume?

OpenStudy (anonymous):

letter b the answer if a and b are constant

OpenStudy (anonymous):

yes your correct eseidl

OpenStudy (anonymous):

and a and b are constants

OpenStudy (anonymous):

but everytime i put the answer its like common that i cant get the answer..

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OpenStudy (anonymous):

Stephanie is correct, the answer is b

OpenStudy (anonymous):

omg thanks

OpenStudy (anonymous):

np

OpenStudy (anonymous):

do you want to see the solution?

OpenStudy (anonymous):

can u check this 1 im sure im correct http://openstudy.com/study?login#/updates/4f5cd533e4b0602be4386441

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