Partial Derivatives of Arctan: A Quick Guide

[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

 

[quasar987]

Mmmh. Did you know that

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

?

 

[James R]

Start with fact that

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

then apply the chain rule.

Does that help?

 

[arildno]

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

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

 

[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}}

 

[tiagobt]

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}}

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)

 

[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?

 

[neutrino]

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?

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

 

[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?

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.

 

[whozum]

y = arctanx [/itex]<br /> <br /> tan y = x [/itex]&lt;br /&gt; &lt;br /&gt; Implicit differentiation wrt x:&lt;br /&gt; &lt;br /&gt; sec^2y \frac{dy}{dx} = 1&lt;br /&gt; &lt;br /&gt; Solve that for dy/dx. Draw a triangle, angle &amp;#039;y&amp;#039;, opposite &amp;#039;x&amp;#039; and adjacent &amp;#039;1&amp;#039;. Find the hypotenuse, and describe the resulting expression in terms of x.&lt;br /&gt; &lt;br /&gt; edit: wow I am slow

 

[tiagobt]

I'm going to 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&#039; \left(\frac x y\right) . \frac 1 y = \frac 1 y
\arctan&#039; \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

 

[whozum]

I'm sorry that's 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.

 

[tiagobt]

I'm sorry that's 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.

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

 

[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.

 

[tiagobt]

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.

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.

 

[whozum]

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