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How do you find the derivative of #cosx/(sinx-2)#?

1 Answer

Mar 30, 2015

The answer is: #y'=(2sinx-1)/(sinx-2)^2#

This is because:

#y'=(-sinx*(sinx-2)-cosx*cosx)/(sinx-2)^2=#

#=(-sin^2x+2sinx-cos^2x)/(sinx-2)^2=(2sinx-1)/(sinx-2)^2#

This is because: #sin^2x+cos^2x=1#.

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