How do you differentiate #tan 2x = cos 3y#?

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Answer 1 · 2015-10-14T00:31:03

#-(2sec^2(2x)*csc(3y))/3 = dy/dx#

Using the chain rule, where #u = 2x# and #d/dx = d/(du)*(du)/dx#

#d/(du)(tan(u))*(du)/dx = d/dx(cos(3y))#

#sec^2(u)*2 = d/dx(cos(3y))#

#2sec^2(2x) = d/dx(cos(3y))#

Using the chain rule, where #v = 3y# and #d/dx = d/(dv)*(dv)/dy*dy/dx#

#2sec^2(2x) = d/(dv)(cos(v))*d/dy(3y)*dy/dx#

#2sec^2(2x) = -3sin(v)*dy/dx#

#-(2sec^2(2x)*csc(3y))/3 = dy/dx#