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

1 Answer

Jul 18, 2017

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)#

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