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

1 Answer
Sep 11, 2014

Most people remember this
#f'(x)=1/{sqrt{1-x^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{1-sin^2y}#,
#{dy}/{dx}=1/sqrt{1-sin^2y}#

By #siny=x#,
#{dy}/{dx}=1/sqrt{1-x^2}#