What is the derivative of #y=arctan(x)#?

1 Answer
Wataru
Aug 30, 2014

The derivative of #y=arctan x# is #y'=1/{1+x^2}#.

We can derive this by using implicit differentiation.

Since inverse tangent is hard to deal with, we rewrite it as
#tan(y) =x#

By implicitly differentiating with respect to #x#,
#sec^2(y)cdot y'=1#

By solving for #y'# and using #sec^2(y)=1+tan^2(y)#,
#y'=1/{sec^2(y)}=1/{1+tan^2(y)}#

Hence, #y'=1/{1+x^2}#.

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