The partial derivatives of arctan(y/x)

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[SOLVED] The partial derivatives of arctan(y/x)

let w = arctan(y/x)

the partial derivatives are:
dw/dx and dw/dy

i know that the derivative or arctan(x) is 1/(1+x^2).
so for dw/dy, i get (1/ 1 + (y^2/x^2) ) * (1/x) = x/(x^2 + y^2) ? correct?

how do i find dw/dx?
 
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That looks right. To get the other, just take d/du(arctan(u)) * du/dx, where u = y/x, just like in the previous situation.
 
The partials are:

dw/dy = 1/ (x + y^2/x )

&

dw/dx = y/( 1+(y^2/x^2)

The total integral, then, is as follows.

w = arctan(y/x)
dw= d(arctan(y/x) d(y/x)
dw= 1/(1+(y/x)^2) (ydx-xdy)/(x^2)
dw= (ydx-xdy)/[x^2(1+(y/x)^2)]
dw= (ydx-xdy)/(x^2+y^2)
 


although it has passed a lot of time, your answer has helped me i thank you for it, but i have something to say: i think you have written something in the uncorrect order, that's 'dw= (ydx-xdy)/(x^2+y^2)' , because if the function is y/x, the rule of the chain would be: dy*x-dx*y, then the final result (from my point of view) is:

dw= (dy*x-x*dy)/(x^2+y^2) =-y/(x^2+y^2).

please if I'm wrong, simply tell me.
 


BrendanH said:
The partials are:

dw/dy = 1/ (x + y^2/x )

&

dw/dx = y/( 1+(y^2/x^2)

The total integral, then, is as follows.

w = arctan(y/x)
dw= d(arctan(y/x) d(y/x)
dw= 1/(1+(y/x)^2) (ydx-xdy)/(x^2)
dw= (ydx-xdy)/[x^2(1+(y/x)^2)]
dw= (ydx-xdy)/(x^2+y^2)

please check my answer above
 


And what would be the time derivative of atan(y/x), where y and x are both functions of time. In other words, what is d(theta)/dt, the time derivative of the spherical coordinate theta?
 


Fisrt, let the function be W =arctan(y/x)
Also, let W=arctan(U), such that U=y/x,
then,
dW= (dW/dU)(dU/dx) + (dW/dU)(dU/dy)...1
But,
dW/dU= 1/(1+U**2).........2
Also,
dU/dx= -y(x)**-2, and

dU/dy= 1/x

.: dW= (1/(1+U**2))(-y(x)**-2) + (1/(1+U**2))(1/x)...3

Putting the values U in Equation 3,

dW= (1/(1+(y/x)**2)(-y(x)**-2 + (1/(1+(y/x)**2)(1/x)

= (1/(1+(y/x)**2)((-y/x) + (1/x)) or

((1/(1+(y/x)**2)((x-y)/(x)**2)).
 

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