How do you find dy/dx by implicit differentiation for #sinx + 2cos2y = 1#?

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Answer 1 · 2015-08-30T15:23:36

#dy/dx=cos(x)/(4sin(2y))#

Differentiate every term with respect to #x#

#cos(x)-4sin(2y)dy/dx=0#

isolate the term containing #dy/dx# by adding

#4sin(2y)dy/dx# to both sides

#4sin(2y)dy/dx=cos(x)#

Divide both sides by #4sin(2y)#

#dy/dx=cos(x)/(4sin(2y))#