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

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Answer 1 · 2015-04-13T17:07:15

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

#y=tan^-1(x)#

#tany=x#

Differentiating both sides with respect to 'x'

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

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

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

So,

# (dy)/(dx)=1/(1+tan^2y)#

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