# How do you find the derivative of y = arctan [ tanx / (sqrt(2+tan^2(x))) ] + ln [tanx + sqrt(2+tan^2(x)) ]?

Mar 29, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{2 + {\tan}^{2} x}$
The solution is to long to show all steps,with formulae and simplifiction.

#### Explanation:

Let ,$y = u + v , w h e r e ,$

$u = {\tan}^{-} 1 \left[\tan \frac{x}{\sqrt{2 + {\tan}^{2} x}}\right] \mathmr{and} v = \ln | \tan x + \sqrt{2 + {\tan}^{2} x} |$

• $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {\left(\tan \frac{x}{\sqrt{2 + {\tan}^{2} x}}\right)}^{2}} \times \frac{d}{\mathrm{dx}} \left(\tan \frac{x}{\sqrt{2 + {\tan}^{2} x}}\right)$
=cancel((2+tan^2x))/(2+tan^2x+tan^2x)xx(((sqrt(2+tan^2x))(sec^2x)-tanx((cancel(2)tanxsec^2x)/(cancel(2)sqrt(2+tan^2x))))/(cancel((2+tan^2x)))
$= \left(\frac{{\sec}^{2} x \left(\sqrt{2 + {\tan}^{2} x} - {\tan}^{2} \frac{x}{\sqrt{2 + {\tan}^{2} x}}\right)}{2 + 2 {\tan}^{2} x}\right)$
$= \frac{{\sec}^{2} x \left(2 + {\tan}^{2} x - {\tan}^{2} x\right)}{2 \left({\sec}^{2} x\right) \left(\sqrt{2 + {\tan}^{2} x}\right)} = \frac{1}{\sqrt{2 + {\tan}^{2} x}} \ldots \ldots \ldots . \left(I\right)$
• $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{\tan x + \sqrt{2 + {\tan}^{2} x}} \left({\sec}^{2} x + \frac{2 \tan x {\sec}^{2} x}{2 \sqrt{2 + {\tan}^{2} x}}\right)$
$= {\sec}^{2} \frac{x}{\tan x + \sqrt{2 + {\tan}^{2} x}} \left(1 + \tan \frac{x}{\sqrt{2 + {\tan}^{2} x}}\right)$
$= {\sec}^{2} \frac{x}{\tan x + \sqrt{2 + {\tan}^{2} x}} \left(\frac{\sqrt{2 + {\tan}^{2} x} + \tan x}{\sqrt{2 + {\tan}^{2} x}}\right)$
$= \frac{1 + {\tan}^{2} x}{\sqrt{2 + {\tan}^{2} x}} \ldots \ldots \ldots . \left(I I\right)$
Using $\left(I\right) \mathmr{and} \left(I I\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} + \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{\sqrt{2 + {\tan}^{2} x}} + \frac{1 + {\tan}^{2} x}{\sqrt{2 + {\tan}^{2} x}} = \frac{2 + {\tan}^{2} x}{\sqrt{2 + {\tan}^{2} x}} = \sqrt{2 + {\tan}^{2} x}$