What is the derivative of #arcsin((2x)/(1+x^2))#?

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Answer 1 · 2016-07-07T09:29:41

#2/(1+x^2)#

' means differentiation with respect to x.

Let #x=tan(theta/2)#. Then, #theta=2tan^(-1)x and theta'=2/(1+x^2)#.

Now, #(arc sin ( (2x)/(1+x^2) )'#

#=(arc sin ( ( 2 tan (theta / 2) )/( 1+tan^2(theta/2))))'#

#=(arc sin (sin theta))'#

#=theta'#

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