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How do you differentiate #y=x/(2-tanx)#?

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Mar 16, 2018

#(2-tanx+xsec^2x)/(2-tanx)^2#

Explanation:

#y=x/(2-tanx)#
#color(red)(d/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/v^2)#
So,
#(dy)/(dx)=((2-tanx)*1-x(-sec^2x))/(2-tanx)^2=(2-tanx+xsec^2x)/(2-tanx)^2#

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