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

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