How can you differentiate #sin^-1((2x)/(1+x^2))# ?

1 Answer
Jul 29, 2018

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

Explanation:

Let ,

#y=sin^-1((2x)/(1+x^2))=arcsin((2x)/(1+x^2))#

Let ,

#y=arcsinu # , where , #color(green)(u=(2x)/(1+x^2)#

So, #(dy)/(du)=1/sqrt(1-u^2) and #

#(du)/(dx)=((1+x^2)*2-2x(0+2x))/(1+x^2)^2 to[because "quotient -rule"]#

#:.(du)/(dx)=(2+2x^2-4x^2)/(1+x^2)^2=(2(1-x^2))/(1+x^2)^2#

#"Using " color(blue) "Chain Rule :"#

#color(blue)((dy)/(dx)=(dy)/(du)(du)/(dx#

#:.(dy)/(dx)=1/sqrt(1-u^2)xx(2(1-x^2))/(1+x^2)^2#

Subst. #color(green)(u=(2x)/(1+x^2)# ,we get

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

#=(1+x^2)/sqrt((1+x^2)^2-4x^2) xx(2(1-x^2))/(1+x^2)^2#

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

#=(1+x^2)/|1-x^2| xx(2(1-x^2))/(1+x^2)^2to[becausesqrt((a)^2)=|a|#

#=(2(1-x^2))/(|1-x^2|(1+x^2))#