Differentiating Inverse Trigonometric Functions
Key Questions

By Implicit Differentiation,
#y'=1/sqrt{1x^2}# .Let us look at some details.
#y=cos^{1}x# by rewriting in term of cosine,
#Rightarrow cos y=x# by implicitly differentiating with respect to
#x# ,
#Rightarrow sin y cdot y'=1# by dividing by
#sin y# ,
#y'=1/sin y# by the trionometric identity
#sin y =sqrt{1cos^2y}# ,
#y'=1/sqrt{1cos^2y}# by
#cos y =x# ,
#y'=1/sqrt{1x^2}# 
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)# 
The answer is
#y'=1/(1+x^2)# We start by using implicit differentiation:
#y=cot^(1)x#
#cot y=x#
#csc^2y (dy)/(dx)=1#
#(dy)/(dx)=1/(csc^2y)#
#(dy)/(dx)=1/(1+cot^2y)# using trig identity:#1+cot^2 theta=csc^2 theta#
#(dy)/(dx)=1/(1+x^2)# using line 2:#cot y = x# The trick for this derivative is to use an identity that allows you to substitute
#x# back in for#y# because you don't want leave the derivative as an implicit function; substituting#x# back in will make the derivative an explicit function. 
Most people remember this
#f'(x)=1/{sqrt{1x^2}}#
as one of derivative formulas; however, you can derive it by implicit differentiation.Let us derive the derivative.
Let#y=sin^{1}x# .By rewriting in terms of sine,
#siny=x# By implicitly differentiating with respect to
#x# ,
#cosy cdot {dy}/{dx}=1# By dividing by
#cosy# ,
#{dy}/{dx}=1/cosy# By
#cosy=sqrt{1sin^2y}# ,
#{dy}/{dx}=1/sqrt{1sin^2y}# By
#siny=x# ,
#{dy}/{dx}=1/sqrt{1x^2}#
Questions
Differentiating Trigonometric Functions

Limits Involving Trigonometric Functions

Intuitive Approach to the derivative of y=sin(x)

Derivative Rules for y=cos(x) and y=tan(x)

Differentiating sin(x) from First Principles

Special Limits Involving sin(x), x, and tan(x)

Graphical Relationship Between sin(x), x, and tan(x), using Radian Measure

Derivatives of y=sec(x), y=cot(x), y= csc(x)

Differentiating Inverse Trigonometric Functions