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

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Answer 1 · 2015-07-05T03:11:15

Assuming you don't remember the derivative of #arctanu#:

#y = arctanx^2#

#tany = x^2#

#sec^2y ((dy)/(dx)) = 2x#

#(dy)/(dx) = (2x)/sec^2y#

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

Since #tany = x^2#:

#= color(blue)((2x)/(1+x^4)#

If you did remember it:

#d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))#

#u = x^2#

#=> 1/(1+(x^2)^2)*2x = color(darkblue)((2x)/(1+x^4))#