# What is the derivative of y = arctan(x - sqrt(1+x^2))?

Aug 19, 2017

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

#### Explanation:

The derivative of $\arctan x$ is $\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$, so the chain rule tells us that when we have a function inside the arctangent function, $\frac{d}{\mathrm{dx}} \arctan u = \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$.

Thus:

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

Note that ${\left(x - \sqrt{1 + {x}^{2}}\right)}^{2} = {x}^{2} - 2 x \sqrt{1 + {x}^{2}} + \left(1 + {x}^{2}\right)$.

Also note that $\frac{d}{\mathrm{dx}} \sqrt{1 + {x}^{2}} = \frac{d}{\mathrm{dx}} {\left(1 + {x}^{2}\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \left(2 x\right) = \frac{x}{\sqrt{1 + {x}^{2}}}$.

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

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

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