# 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…

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