derivative of arctan2x^1/2

• November 21st 2011, 02:57 PM
Barthayn
derivative of arctan2x^1/2
[QUOTE=Quacky;694202]Let's see.

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

How do I differentiate from here? I get $\frac {dy}{dx} = \frac {1}{4x^3+x}$ as an answer
• November 21st 2011, 03:04 PM
e^(i*pi)
Re: derivative of arctan2x^1/2
[QUOTE=Barthayn;696842]
Quote:

Originally Posted by Quacky
Let's see.

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

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

Use implicit differentiation and the chain rule.

The derivative of $\tan(y^2)$ is $2y \sec^2(y^2) \cdot \dfrac{dy}{dx}$
• November 21st 2011, 03:07 PM
skeeter
Re: derivative of arctan2x^1/2
let $u = 2x$

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

$y' = \frac{1}{2}\left(\arctan{u}\right)^{-1/2} \cdot \frac{u'}{1+u^2}$
• November 21st 2011, 03:16 PM
Barthayn
Re: derivative of arctan2x^1/2
thanks, I got it now :)