# How do you differentiate arctan(x-(1+x^2)^(1/2))?

Jun 6, 2017

$= \left[\frac{1}{\sqrt{{\left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2} + 1}}\right] \left[1 - x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}\right]$

#### Explanation:

Before answering let us differentiate the inverse tan function.

$\arctan \left(x\right) = y$
$\tan \left(\arctan \left(x\right)\right) = \tan \left(y\right)$
$\frac{d}{\mathrm{dx}} \left[x\right] = \frac{d}{\mathrm{dx}} \left[\tan \left(y\right)\right]$
$1 = {\sec}^{2} \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{1}{{\sec}^{2} \left(\arctan x\right)} = \frac{\mathrm{dy}}{\mathrm{dx}}$

But $\sqrt{{\tan}^{2} \left(\alpha\right) + 1} = {\sec}^{2} \left(\alpha\right)$

Then in which $\alpha = \arctan \left(x\right)$:
$\frac{1}{\sqrt{{\tan}^{2} \left(\arctan \left(x\right)\right) + 1}} = \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{1}{\sqrt{{x}^{2} + 1}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Now differentiating:
$\frac{d}{\mathrm{dx}} \left[\arctan \left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)\right]$
$= \left[\frac{1}{\sqrt{{\left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2} + 1}}\right] \frac{d}{\mathrm{dx}} \left[x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right]$
$= \left[\frac{1}{\sqrt{{\left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2} + 1}}\right] \left[1 - \left(\frac{1}{2}\right) {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left[{x}^{2} + 1\right]\right]$
$= \left[\frac{1}{\sqrt{{\left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2} + 1}}\right] \left[1 - \frac{{\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}}{2} \left[2 x\right]\right]$
$= \left[\frac{1}{\sqrt{{\left(x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)}^{2} + 1}}\right] \left[1 - x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}\right]$