# derivative of arctan2x^1/2

• Nov 21st 2011, 01:57 PM
Barthayn
derivative of arctan2x^1/2
[QUOTE=Quacky;694202]Let's see.

$\displaystyle y = \sqrt{arctan2x}$
$\displaystyle y^2 = arctan2x$
$\displaystyle tany^2 = 2x$
$\displaystyle \frac {dy}{dx} * sec^2y^2$ $\displaystyle tany^2 = 2$

How do I differentiate from here? I get $\displaystyle \frac {dy}{dx} = \frac {1}{4x^3+x}$ as an answer
• Nov 21st 2011, 02:04 PM
e^(i*pi)
Re: derivative of arctan2x^1/2
[QUOTE=Barthayn;696842]
Quote:

Originally Posted by Quacky
Let's see.

$\displaystyle y = \sqrt{arctan2x}$
$\displaystyle y^2 = arctan2x$
$\displaystyle tany^2 = 2x$
$\displaystyle \frac {dy}{dx} * sec^2y^2$ $\displaystyle tany^2 = 2$

How do I differentiate from here? I am lost because of the $\displaystyle y^2$

Use implicit differentiation and the chain rule.

The derivative of $\displaystyle \tan(y^2)$ is $\displaystyle 2y \sec^2(y^2) \cdot \dfrac{dy}{dx}$
• Nov 21st 2011, 02:07 PM
skeeter
Re: derivative of arctan2x^1/2
let $\displaystyle u = 2x$

$\displaystyle y = \left(\arctan{u}\right)^{1/2}$

$\displaystyle y' = \frac{1}{2}\left(\arctan{u}\right)^{-1/2} \cdot \frac{u'}{1+u^2}$
• Nov 21st 2011, 02:16 PM
Barthayn
Re: derivative of arctan2x^1/2
thanks, I got it now :)