# Partial derivatives of arccos

• Jun 18th 2013, 12:46 PM
Vulpes
Partial derivatives of arccos
Hi All,

To be able to construct a Jacobian, I am asked to find the two partial derivatives of the following inverse trigonometric function:

$w_{1}(u,v) = \frac{1}{2\pi} \arccos (\frac{u}{\sqrt{u^{2} + v^{2}}})$

I know that $\frac{\delta \arccos x}{\delta x} = \frac{-1}{\sqrt{1-x^2}}$, but I am realy stumped at how to proceed from this to get the partial derivatives.

The anwers to the partian derivatives are given as:

$\frac{\delta w_{1}}{\delta u} = \frac{-v}{2\pi(u^{2}+v^{2})}$

$\frac{\delta w_{1}}{\delta v} = \frac{u}{2\pi(u^{2}+v^{2})}$

Any help on how to arrive here would be much appreciated!
Kind regards,
Chris
• Jun 18th 2013, 01:43 PM
HallsofIvy
Re: Partial derivatives of arccos
Use the chain rule. If $w_1= \frac{1}{2\pi}arcccos(\frac{u}{\sqrt{u^2+ v^2}}$, then, yes, the derivative of arcos(x) is $\frac{-1}{\sqrt{1- x^2}}$.
Let $x(u,v)= \frac{u}{\sqrt{u^2+ v^}}$ so that $w_1(u)= \frac{1}{2\pi}arcos(x)$
Now, the chain rule says that $\frac{\partial w_1}{\partial u}= \frac{dw_1}{dx}\frac{\partial x}{\partial u}$
You already know that $\frac{dw_1}{dx}= \frac{-1}{\sqrt{1- x^2}}= \frac{-1}{\sqrt{1- \frac{u^2}{u^2+ v^2}}}$
and now you need to multiply that by $\frac{\partial x}{\partial u}= \frac{\partial}{\partial u}\left(\frac{u}{\sqrt{u^2+ v^2}}\right)$