What is the derivative of #y=tan(x+y)#?

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Answer 1 · 2016-07-18T13:20:12

#:. dy/dx=-(1+y^2)/y^2#, or,

#=-csc^2(x+y)#

#y=tan(x+y) rArr tan^-1y=x+y rArr tan^-1y-y=x#

Diff.ing w.r.t. #y#, we have,

#1/(1+y^2)-1=dx/dy#

#:. (1-1-y^2)/(1+y^2)=dx/dy#

#:.-y^2/(1+y^2)=dx/dy#

#:. dy/dx=1/(dx/dy)=-(1+y^2)/y^2#, or,

#dy/dx=-{1+tan^2(x+y)}/tan^2(x+y)=-sec^2(x+y)/tan^2(x+y)=-csc^2(x+y)##