# What's the derivative of arctan(x - sqrt(1+x^2))?

Mar 29, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{2 \left(1 + {x}^{2}\right)}$

#### Explanation:

Let us first workout derivative of $f \left(u\right) = \arctan u$.

As $g \left(u\right) = \arctan u$, $\tan g = u$ and differentiating we have

${\sec}^{2} g \cdot \frac{\mathrm{dg}}{\mathrm{dx}} = 1$ or $\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} g} = \frac{1}{1 + {\tan}^{2} g} = \frac{1}{1 + {u}^{2}}$

Hence for $g \left(u\right) = \arctan u$, we have (dg)/(dx)=1/(1+u^2....(A)

Now to differentiate $f \left(x\right) = \arctan \left(x - \sqrt{1 + {x}^{2}}\right)$, we will use (A) and chain rule.

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(x - \sqrt{1 + {x}^{2}}\right)}^{2}}\right) \cdot \left(1 - \frac{1}{2 \sqrt{1 + {x}^{2}}} \cdot 2 x\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(x - \sqrt{1 + {x}^{2}}\right)}^{2}}\right) \cdot \left(1 - \frac{x}{\sqrt{1 + {x}^{2}}}\right)$

If $x = \tan \theta$

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(\tan \theta - \sqrt{1 + {\tan}^{2} \theta}\right)}^{2}}\right) \cdot \left(1 - \tan \frac{\theta}{\sqrt{1 + {\tan}^{2} \theta}}\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\left(\tan \theta - \sec \theta\right)}^{2}}\right) \cdot \left(1 - \tan \frac{\theta}{\sec} \theta\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{1 + {\tan}^{2} \theta + {\sec}^{2} \theta - 2 \tan \theta \sec \theta}\right) \cdot \left(\frac{\sec \theta - \tan \theta}{\sec} \theta\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{2 \sec \theta \left(\sec \theta - \tan \theta\right)}\right) \cdot \left(\frac{\sec \theta - \tan \theta}{\sec} \theta\right)$ or

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(\frac{1}{2 {\sec}^{2} \theta}\right) = \frac{1}{2 \left(1 + {\tan}^{2} \theta\right)} = \frac{1}{2 \left(1 + {x}^{2}\right)}$