# 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…

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

# 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:

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

# How to differentiate y=arctanx

Below I’m going to demonstrate how to integrate y=arctanx…

Firstly, we need to know that:

$\sin ^{ 2 }{ y } +\cos ^{ 2 }{ y } =1\\ \\ \frac { \sin ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } +\frac { \cos ^{ 2 }{ y } }{ \cos ^{ 2 }{ y } } =\frac { 1 }{ \cos ^{ 2 }{ y } } \\ \\ \tan ^{ 2 }{ y } +1=\sec ^{ 2 }{ y }$

We also need to know that:

$x=\tan { y } \\ \\ x=\frac { \sin { y } }{ \cos { y } } \\ \\ x\cdot \cos { y } =\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } +x\cdot \left( -\sin { y } \right) =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } -x\sin { y } =\cos { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +x\sin { y } \\ \\ \frac { dx }{ dy } \cdot \cos { y } =\cos { y } +\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ \frac { 1 }{ \cos { y } } \cdot \frac { dx }{ dy } \cdot \cos { y } =\frac { 1 }{ \cos { y } } \left( \cos { y+\frac { \sin ^{ 2 }{ y } }{ \cos { y } } } \right) \\ \\ \frac { dx }{ dy } =1+\tan ^{ 2 }{ y } \\ \\ \therefore \quad \frac { dx }{ dy } =\sec ^{ 2 }{ y }$

And finally:

$\frac { dy }{ dy } \cdot \frac { dy }{ dx } =\frac { dy }{ dx }$

Now, using implicit differentiation:

$y=\arctan { x } \\ \\ \tan { y } =x\\ \\ \sec ^{ 2 }{ y } \cdot \frac { dy }{ dx } =1\\ \\ \frac { dy }{ dx } =\frac { 1 }{ \sec ^{ 2 }{ y } } \\ \\ \frac { dy }{ dx } =\frac { 1 }{ \tan ^{ 2 }{ y+1 } } \\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 1 }{ { x }^{ 2 }+1 }$