# Partial derivatites of arctan

1. May 16, 2005

### tiagobt

Partial derivatives of arctan

Could anyone help me with the following partial derivatives?

$$\nabla \arctan \left(\frac x y \right)$$

$$\nabla \arctan \left(\frac y x \right)$$

Thanks,

Tiago

Last edited: May 16, 2005
2. May 17, 2005

### quasar987

Mmmh. Did you know that

$$\frac{d}{dx}Arctg[f(x)] = \frac{f'(x)}{1+[f(x)]^2}$$

?

3. May 17, 2005

### James R

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

then apply the chain rule.

Does that help?

4. May 17, 2005

### arildno

It is of some interest to note that for any "x", we have the equality:
$$arctan(x)+arctan(\frac{1}{x})=\frac{\pi}{2}$$

5. May 17, 2005

### whozum

Just to clear up notation issues, he's looking for the partial derivative for each variable right? Such as

$$\nabla_x = \frac{1}{y+\frac{x^2}{y}}$$

6. May 17, 2005

### tiagobt

For example:

$$\nabla \arctan \left(\frac x y \right) = \left(\frac \partial {\partial x} \arctan \left(\frac x y \right), \frac \partial {\partial y} \arctan \left(\frac x y \right)\right)$$

7. May 17, 2005

### tiagobt

Actually, I don't even remember how to find the derivative of $$\arctan(x)$$ (the formulas you posted). I tried something like this:

$$\tan \left(\arctan(x) \right) = x$$

Differentiating in respect to $$x$$:

$$\sec ^2 \left(\arctan(x) \right).\arctan ' (x) = 1$$
$$\arctan '(x) = \frac 1 {\sec ^2 \left(\arctan(x) \right)}$$

Not really what I was supposed to find... Could you please give me any more hints? :uhh:

8. May 17, 2005

### neutrino

Now rewrite the RHS of the last equation in terms of tan using a simple trigonometric identity.

9. May 17, 2005

### tiagobt

OK, I figured it out myself.

Now I use:

$$\sec^2 \left(\arctan(x) \right) = \tan^2 \left(\arctan (x) \right) + 1 = x^2 + 1$$

What gives me:

$$\arctan '(x) = \frac 1 {x^2 + 1}$$

Maybe I can do something similar with the original derivatives. I'll give it a try.

10. May 17, 2005

### whozum

$$y = arctanx [/itex] [tex] tan y = x [/itex] Implicit differentiation wrt x: [tex] sec^2y \frac{dy}{dx} = 1$$

Solve that for dy/dx. Draw a triangle, angle 'y', opposite 'x' and adjacent '1'. Find the hypotenuse, and describe the resulting expression in terms of x.

edit: wow im slow

11. May 17, 2005

### tiagobt

I'm gonna try:

$$\frac \partial {\partial x} \arctan \left(\frac x y \right)$$

$$\tan \left(\arctan \left(\frac x y\right) \right) = \frac x y$$

Derivative with respect to x:

$$\sec^ 2 \left(\arctan \left(\frac x y\right) \right).\arctan' \left(\frac x y\right) . \frac 1 y = \frac 1 y$$
$$\arctan' \left(\frac x y\right) \right = \frac 1 {\frac {x^2} {y^2} + 1} = \frac {y^2} {x^2 + y^2}$$

Is this right?

Thanks again

Last edited: May 17, 2005
12. May 17, 2005

### whozum

I'm sorry thats really hard to follow, name f(x,y) = z and try going through again, for me. Your result is not the correct answer though, it looks like you forgot to apply the chain rule.

13. May 17, 2005

### tiagobt

Do you want me to name $$f (x,y) = \arctan \left(\frac x y \right)$$ ? I'm not sure what you mean.

14. May 17, 2005

### dextercioby

Nope,it's incorrect.

$$\nabla \arctan\frac{x}{y}=\frac{\partial \arctan\frac{x}{y}}{\partial x}\vec{i}+\frac{\partial \arctan\frac{x}{y}}{\partial y} \vec{j}$$

Take the first derivative

$$\frac{\partial \arctan\frac{x}{y}}{\partial x}=\frac{1}{y}\frac{1}{1+\frac{x^{2}}{y^{2}}}=\frac{y}{x^{2}+y^{2}}$$

You do the second.

Daniel.

15. May 17, 2005

### tiagobt

I think I got it now:

$$\frac{\partial \arctan\frac{x}{y}}{\partial y}=\frac{-x}{y^2}\frac{1}{\frac{x^{2}}{y^{2}}+1}=-\frac{x}{x^{2}+y^{2}}$$

I just can't figure out what I did wrong in the other post.

But that's OK. Thanks a lot.

16. May 17, 2005

### whozum

You didnt apply the chain rule the first time, which you did this time. So you are fine.