How do you differentiate # arctan(x^2+1)#?

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Answer 1 · 2016-06-22T08:16:11

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

let # y = arctan (x^2 +1)# so #tan y = x^2 + 1#

thus #sec^2 y \ y' = 2x, \qquad y' = (2x)/(sec^2 y)#

using identity #tan^2 + 1 = sec^2# we have

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

Answer 2 · 2016-06-22T08:16:27

#dy/dx=(2x)/(x^4+2x^2+2).#

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

Then, #dy/dx=d/dx{arctan(x^2+1)} =1/{1+(x^2+1)^2}*d/dx(x^2+1)#................[Chain rule]
#=1/(1+x^4+2x^2+1)*2x=(2x)/(x^4+2x^2+2).#