How do you differentiate $y=sin(arccost)$ ?

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Answer 1 · 2017-07-18T09:41:24

We will use the chain rule.

$y=sin(arccost)$

so

$dy/dt=cos(arccost)*(d(arccost))/dt=$

$cos(arccost)(-1/sqrt(1-t^2))=-cos(arccost)/sqrt(1-t^2)$

so

$dy/dt=-cos(arccost)/sqrt(1-t^2)$

How do you differentiate y=sin(arccost)y=sin(arccost)?