CALCULUS DERIVATIVE HELP! y=arcsin(costheta)?

Question

I simplified the equation down to -sintheta/sq rt 1-costheta then down to -sintheta/sintheta then down to -1. Now, I know my derivative is correct, but my instructor said it could be considered a piecewise function. Can someone explain where I went wrong in simplifying?

? · Accepted Answer

y = arc sin(cos theta)
sin y = cos theta
Now take the derivative wrt theta
cos y dy/dtheta = - sin theta
This gives
d y/ d theta  = - sin theta/cosy = -sin theta /sqrt(1-sin^2 y)=- sin theta/(sqrt( 1- cos^2 theta)

Anonymous · Answer

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CALCULUS DERIVATIVE HELP! y=arcsin(costheta)?

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