Question

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

Answer 1

We will use the chain rule.

Explanation:

`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)`