How do you differentiate #sec(arctan(x))#?

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Answer 1 · 2016-08-20T11:29:48

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

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if #tan theta = x# then #theta = arctan x#

and #sec (arctan x ) = sec theta#

#sec theta = 1/(cos theta) = 1/( (1)/(sqrt(1+x^2))) = sqrt (1+ x^2)#

it follows that

#d/dx ( sec (arctan (x)) #

#= d/dx sqrt(1+ x^2) = #

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