How do you find the derivative of #y= 1 / (2 sin 2x)#?

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Answer 1 · 2018-04-13T07:43:50

The derivative is #=-(cos2x)/(sin^2 2x)#

We need

#(1/(u(x)))'=-(u'(x))/(u(x))^2#

Here,

#y=1/(2sin(2x))#

#dy/dx=1/2*(1/sin(2x))'#

#=1/2*(-1/(sin^2 2x))*2cos2x#

#=-(cos2x)/(sin^2 2x)#