How do you find the derivative of #arcsin((2x)/(1+x^2))#?

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Answer 1 · 2017-01-24T19:44:13

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

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

#siny=(2x)/(1+x^2)#

#dy/dx=1/(dx/dy)#

#dx/dy=cosy#

#dy/dx=1/cosy#

#sin^2y+cos^2y=1#

#cos^2y=1-sin^2y#

#cosy=sqrt(1-sin^2y)=sqrt(1-(4x^2)/(x^2+1)^2#

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