$\text{find the gradient field}$ $\vec{F}=\vec{V}\varphi$
$\text{for the following functions$\varphi$}\\$
$\text{a.$\displaystyle\vec{V}(x,y)=\tan^{-1}\left(\frac{y}{x} \right)$}\\$
$\text{b.$\displaystyle\vec{V}(x,y,z)=\cos\left(\frac{y}{x} \right)$}$

ok this completely new to me

did you learn what the gradient is?

The gradient of a scalar function $f(x,y,z)$ is known as $\nabla f$ and in cartesian coordinates $\nabla f = \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)$

Since you have only two independent variables, x and y, you must have $\displaystyle \nabla f= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.
(I suspect that what you have written as "$\displaystyle \vec{V}f$" was intended to be $\displaystyle \nabla f$".)

So what are $\displaystyle \frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial x}$ and $\displaystyle \frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}$?

And what are $\displaystyle \frac{\partial cos\left(\frac{y}{x}\right)}{\partial x}$ and $\displaystyle \frac{\partial cos\left(\frac{y}{x}\right)}{\partial y}$?

Originally Posted by HallsofIvy
Since you have only two independent variables, x and y, you must have $\displaystyle \nabla f= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.
(I suspect that what you have written as "$\displaystyle \vec{V}f$" was intended to be $\displaystyle \nabla f$".)

So what are $\displaystyle \frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial x}$ and $\displaystyle \frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}$?

And what are $\displaystyle \frac{\partial cos\left(\frac{y}{x}\right)}{\partial x}$ and $\displaystyle \frac{\partial cos\left(\frac{y}{x}\right)}{\partial y}$?
do you mean this?
$\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}=-\dfrac{\arctan\left(y\right)}{x^2}$

Originally Posted by bigwave
do you mean this?
$\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}=-\dfrac{\arctan\left(y\right)}{x^2}$
How did you arrive at this? Have you learned how to take partial derivatives?

$\displaystyle \dfrac{\partial \tan^{-1}\left(\dfrac{y}{x}\right)}{\partial y} = \dfrac{1}{1+\left(\dfrac{y}{x}\right)^2}\cdot \dfrac{1}{x} = \dfrac{x}{x^2+y^2}$

no I did it thru an online derivative calculator

Originally Posted by bigwave
no I did it thru an online derivative calculator
Wolfram|Alpha: Computational Knowledge Engine

But, honestly, if you have not learned how to take partial derivatives, this will be next to impossible for you to do on your own. It may be prudent to learn how to take partial derivatives.

I was hoping an example of that would be done here

Originally Posted by bigwave
I was hoping an example of that would be done here
A partial derivative with respect to a variable means treat all other variables as constants and take the derivative. So, in the example $\displaystyle \dfrac{\partial \tan^{-1}\left( \dfrac{y}{x} \right)}{\partial y}$, we are treating $\displaystyle x$ as a constant and then taking the derivative. Do you know how to take derivatives in general? Because there is an entire course on derivatives called Calculus. An example here would not help. You would need to actually learn differential calculus to be able to understand derivatives in general.

Im in calculus III right now
I missed a lot of classes lately so this partial derivaties is new
I have another forum to go to if you don't want to show the steps
I just cant find a good example of this.

$\displaystyle \dfrac{d}{dx}(\tan^{-1} u) = \dfrac{1}{1+u^2}\dfrac{du}{dx}$