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

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Answer 1 · 2017-05-13T14:26:56

The derivative is #=-(2x)/(x^4+2x^2+2)#

We need

#(tanx)'=sec^2x#

#1+tan^2x=sec^2x#

#(1/x)'=-1/x^2#

Let #y=arctan(1/(1+x^2))#

Then,

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

Differentiating both sides

#sec^2y*dy/dx=(-2x)/(1+x^2)^2#

#sec^2y=1+tan^2y=1+(1/(1+x^2))^2=((1+x^2)^2+1)/(1+x^2)^2#

Therefore,

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

#=-(2x)/(x^4+2x^2+2)#