How to differentiate y=arcsinx

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

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

But…

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

Therefore:

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

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