# How do you Find the derivative of y=arctan(x-sqrt(1+x^2))?

Sep 11, 2014

By Chain Rule,
$y ' = \frac{1}{2 \left(1 + {x}^{2}\right)}$

Let us look at some details.

Let us first find the derivative of $x - \sqrt{1 + {x}^{2}}$.
By rewriting the square-root as the 1/2 power,
$\left(x - \sqrt{1 + {x}^{2}}\right) ' = \left[x - {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right] '$

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

by taking the common denominator,
$= \frac{\sqrt{1 + {x}^{2}} - x}{\sqrt{1 + {x}^{2}}}$

Now, we can find $y '$.
By Chain Rule,
$y ' = \frac{1}{1 + {\left(x - \sqrt{1 + {x}^{2}}\right)}^{2}} \cdot \frac{\sqrt{1 + {x}^{2}} - x}{\sqrt{1 + {x}^{2}}}$

by multiplying out the denominator of the first quotient,
$= \frac{1}{2 \left(1 + {x}^{2} - x \sqrt{1 + {x}^{2}}\right)} \cdot \frac{\sqrt{1 + {x}^{2}} - x}{\sqrt{1 + {x}^{2}}}$

by multiply the quotients together,
$= \frac{\sqrt{1 + {x}^{2}} - x}{2 \left[\left(1 + {x}^{2}\right) \sqrt{1 + {x}^{2}} - x \left(1 + {x}^{2}\right)\right]}$

by factoring out $\left(1 + {x}^{2}\right)$ from the denominator,
$= \frac{\sqrt{1 + {x}^{2}} - x}{2 \left(1 + {x}^{2}\right) \left(\sqrt{1 + {x}^{2}} - x\right)}$

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