How do you use implicit differentiation to find dy/dx given #tanx+csc2y=1#?

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Answer 1 · 2017-04-09T11:32:59

#dy/dx=(sec^2 x)/(2csc 2y cot 2y)#

#tan x + csc 2y=1#

By differentiating with respect to #x#,

#sec^2 x-csc 2y cot 2y cdot 2 dy/dx=0#

By subtracting #sec^2 x# from both sides,

#-2csc 2y cot 2y dy/dx=-sec^2 x#

By dividing both sides by #-2csc 2y cot 2y#,

#dy/dx=(sec^2 x)/(2csc 2y cot 2y)#

I hope that this was clear.