How to differentiate y=arccosx

Below I’ll be demonstrating how to differentiate y=arccosx using implicit differentiation…

y=\arccos { x } \\ \\ \cos { y=x } \\ \\ -\sin { y\cdot \frac { dy }{ dx }  } =1\\ \\ \frac { dy }{ dx } =-\frac { 1 }{ \sin { y }  }

But…

\sin ^{ 2 }{ y+\cos ^{ 2 }{ y=1 }  } \\ \\ \sin ^{ 2 }{ y } =1-\cos ^{ 2 }{ y } \\ \\ \sin { y=\sqrt { 1-\cos ^{ 2 }{ y }  }  }

Therefore…

\frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-\cos ^{ 2 }{ y }  }  } \\ \\ \therefore \quad \frac { dy }{ dx } =-\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  }

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