# 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:

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