What is the derivative of #f(x)=tan^-1(x)# ?

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Answer 1 · 2015-07-01T16:47:50

I seem to recall my professor forgetting how to deriving this. This is what I showed him:

#y = arctanx#

#tany = x#

#sec^2y (dy)/(dx) = 1#

#(dy)/(dx) = 1/(sec^2y)#

Since #tany = x/1# and #sqrt(1^2 + x^2) = sqrt(1+x^2)#, #sec^2y = (sqrt(1+x^2)/1)^2 = 1+x^2#

#=> color(blue)((dy)/(dx) = 1/(1+x^2))#

I think he originally intended to do this:

#(dy)/(dx) = 1/(sec^2y)#

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

#tan^2y = x -> sec^2y = 1+x^2#

#=> (dy)/(dx) = 1/(1+x^2)#