Partial Differentiation: How Do You Differentiate Arc Tangent Terms?

[andrew.c]

I was working through this problem,

I understand the method, but got stuck trying to differentiate the arc tan terms...

In the table of standard derivatives,

$$\frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$$

and for PDE's, you treat y as constant when differentiating w.r.t. x

---

Any help greatly appreciated!
:)

 

[Mark44]

I was working through this problem,

I understand the method, but got stuck trying to differentiate the arc tan terms...

In the table of standard derivatives,

$$\frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$$

and for PDE's, you treat y as constant when differentiating w.r.t. x

---

Any help greatly appreciated!
:)

When you are doing partial differentiation, you treat each variable as independent from the others. So when you find the partial of your function w.r.t. x, say, you treat y as a constant.

 

[andrew.c]

I get that - if you could have a look at my attachment, that has the question in it :)

 

[Mark44]

Suppose you had u = u(x, y) = x²tan^-1(y/x), just to make it easier for me to type this in.

$$\frac{\partial u}{\partial x}~=~2x~tan^{-1}(y/x)~+~x^2~\frac{\partial}{\partial x}\left(tan^{-1}(y/x)\right)$$
$$=~2x~tan^{-1}(y/x)~+~x^2~\frac{1}{1 + (y/x)^2}~\frac{\partial}{\partial x}\left(\frac{y}{x}\right)$$
For that last partial, you have what amounts to y*d/dx(1/x^2) = -y/x.

Clear?

 

[andrew.c]

Yeah, I just missed a bit of chain-rule didn't I?
Thanks for the help :)