What is the derivative of #y=arctan sqrt((1-x)/(1+x))#?

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Answer 1 · 2017-11-26T06:52:54

#(dy)/(dx)=-sqrt((1+x)^5)/(2sqrt(1-x))#

Let #u=sqrt((1-x)/(1+x))#. Also observe #(1-x)/(1+x)=1-(2x)/(1+x)#

then using chain rule #(du)/(dx)=1/(2sqrt((1-x)/(1+x)))xxd/(dx)(1-(2x)/(1+x))#

= #sqrt(1+x)/(2sqrt(1-x))xxd/(dx)(1-(2x)/(1+x))#

= #sqrt(1+x)/(2sqrt(1-x))xx(-(2(1+x)-2x)/(1+x)^2)#

= #sqrt(1+x)/(2sqrt(1-x))xx(-2/(1+x)^2)#

= #-sqrt(1+x)^3/sqrt(1-x)#

Hence #y=arctanu# and hence #(dy)/(dx)=1/(1+u^2)xx(du)/(dx)#

i.e. #(dy)/(dx)=1/(1+(1-x)/(1+x))xx(-sqrt((1+x)^3)/sqrt(1-x))#

= #-(1+x)/2xxsqrt((1+x)^3)/sqrt(1-x)#

= #-sqrt((1+x)^5)/(2sqrt(1-x))#