Question

What is the derivative of arctan (y/x) with respect to x??

Answer 1

Inverse tangent:
d/dx [ arctan(u) ] = 1/(u² + 1) · du/dx

Power rule:
d/dx [ 1/u ] = d/dx [ u^(-1) ] = -1/u² · du/dx

Just don't forget the chain rule here:

∂/∂x [ arctan(y/x) ]
= 1/[ (y/x)² + 1 ] ·∂/∂x [ (y/x) ]
= 1/[ (y/x)² + 1 ] · y · ∂/∂x [ (1/x) ]
= 1/[ (y/x)² + 1 ] · y · (-1)/(x²)
= -y / [ x²(y/x)² + 1·x² ]
= -y / [ x²(y²/x²) + x² ]
= -y / ( y² + x² )

Answer:
∂/∂x [ arctan(y/x) ] = -y / ( y² + x² )

Answer 2

thx