How do I find the derivative of $y=arctan(cos θ)$ ?

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Answer 1 · 2015-06-01T17:30:38

$y'=-(sintheta)/(cos^(2)theta+1)$

$y=arctan(costheta)$

Applying the chain rule:

$y'=(1)/(cos^(2)theta+1)xx(d[costheta])/("d"theta)$

$=(-sintheta)/(cos^(2)theta+1)$

$=-(sintheta)/(cos^(2)theta+1)$

How do I find the derivative of y=arctan(cos θ)y=arctan(cos θ)?