Qual é a derivada de # y = arctan sqrt ((1-x) / (1 + x)) #?

Deixei #u=sqrt((1-x)/(1+x))#. Observe também #(1-x)/(1+x)=1-(2x)/(1+x)#

então usando regra da cadeia #(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)#

Conseqüentemente #y=arctanu# e, portanto #(dy)/(dx)=1/(1+u^2)xx(du)/(dx)#

ou seja #(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))#