#x=tan(x+y)#
Diff.ing, both sides w.r.t. #y#, and keeping in mind the Chain Rule,
#dx/dy=d/dytan(x+y)=(sec^2(x+y))d/dy(x+y)=sec^2(x+y)*(dx/dy+1)#
#dx/dy-(sec^2(x+y))dx/dy=sec^2(x+y)#
#{1-sec^2(x+y)}dx/dy=sec^2(x+y)#
Using, #sec^2theta=1+tan^2theta#, we have,
#(-tan^2(x+y))dx/dy=1+tan^2(x+y)#
Knowing that, we have #tan(x+y)=x#.
#-x^2dx/dy=1+x^2#, giving, #dx/dy=-(1+x^2)/x^2#
Therefore, #dy/dx=-x^2/(1+x^2).#
Method II
#x=tan(x+y)#
#rArr d/dxx=d/dx(tan(x+y))#
#rArr 1=sec^2(x+y){d/dx(x+y)}=sec^2(x+y){1+dy/dx}=(1+x^2){1+dy/dx}#
#rArr 1+dy/dx=1/(1+x^2) rArr dy/dx=1/(1+x^2)-1=(1-1-x^2)/(1+x^2)#
#rArr dy/dx=-x^2/(1+x^2),# as in Method I !
Method III
#x=tan(x+y)#
#arctanx=x+y rArr arctanx-x=y#
# rArr dy/dx=1/(1+x^2)-1 =-x^2/(1+x^2)#, as derived before!
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