What is the derivative of #f(x)=arctan[(x+1)/(x-1)]#?

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Answer 1 · 2018-05-14T14:41:26

# rArr f'(x)=-1/(x^2+1), (x!=1)#.

Observe that,

#tan(pi/4+theta)=(tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta)=(1+tantheta)/(1-tantheta)#.

#:. tan(-(pi/4+theta))=-tan(pi/4+theta)=(tantheta+1)/(tantheta-1)#.

Recall that, the range of #tan# function is #RR#.

So, we can very well take #x=tantheta# and get,

#f(x)=arc tan{(x+1)/(x-1)}=arc tan{(tantheta+1)/(tantheta-1)}#,

#=arc tan{tan(-(pi/4+theta)}#,

#=-(pi/4+theta)#.

# rArr f(x)=-pi/4-arc tanx, (x!=1)#.

# rArr f'(x)=0-1/(x^2+1)=-1/(x^2+1), (x!=1)#, is the

desired derivative!