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

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Answer 1 · 2014-09-14T06:59:51

By Implicit Differentiation,
#y'=-1/sqrt{1-x^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{1-cos^2y}#,
#y'=-1/sqrt{1-cos^2y}#

by #cos y =x#,
#y'=-1/sqrt{1-x^2}#