# The partial derivatives of arctan(y/x)

• Hayate
In summary, the partial derivatives of arctan(y/x) are dw/dx = y/( 1+(y^2/x^2) and dw/dy = 1/ (x + y^2/x ) . The total integral is w = arctan(y/x) and the time derivative of arctan(y/x) is d(theta)/dt = (1/(1+(y/x)**2)((x-y)/(x)**2)).
Hayate
[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?

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)

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)).

## 1. What is the definition of partial derivatives of arctan(y/x)?

The partial derivatives of arctan(y/x) are the rate of change of the arctangent function with respect to changes in the variables y and x. It is calculated by taking the derivative of arctan(y/x) with respect to each variable while treating the other variable as a constant.

## 2. How do you find the partial derivatives of arctan(y/x)?

To find the partial derivatives of arctan(y/x), you can use the quotient rule or the chain rule. First, rewrite the function as arctan(yx-1). Then, take the derivative with respect to y or x while treating the other variable as a constant. Repeat for the other variable to find the two partial derivatives.

## 3. What is the significance of the partial derivatives of arctan(y/x)?

The partial derivatives of arctan(y/x) are important in multivariable calculus and optimization problems. They can help to find the maximum or minimum points of a function, as well as the direction of steepest ascent or descent.

## 4. Can the partial derivatives of arctan(y/x) be negative?

Yes, the partial derivatives of arctan(y/x) can be negative. It depends on the values of y and x and the location of the point on the function. A negative partial derivative indicates a decrease in the function's value in that direction, while a positive partial derivative indicates an increase.

## 5. How are the partial derivatives of arctan(y/x) related to the gradient?

The partial derivatives of arctan(y/x) are components of the gradient vector. The gradient is a vector that points in the direction of greatest increase of a function and has a magnitude equal to the rate of change in that direction. The partial derivatives of arctan(y/x) can be used to calculate the gradient at any point on the function.

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