Inverse Trig: Solutions

$y &sp;=&sp; x(sin^-1^x)^2^$
$y' &sp;=&sp; (sin^-1^x)^2^ + 2x(sin^-1^x) 1 - x^2^$
$y &sp;=&sp; tan^-1^(cos x)$
$y' &sp;=&sp; - sin x 1 + cos^2^&sp;=&sp; - sin x$

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