×

Hello! Socratic's Terms of Service and Privacy Policy have been updated, which will be automatically effective on October 6, 2018. Please contact hello@socratic.com with any questions.

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

1 Answer
Jul 1, 2015

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)#