Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 32

Determine the derivative of the function $y = \tan^{-1} \left( x - \sqrt{1 + x^2} \right)$ and simplify if possible.
If $y = \tan^{-1} \left( x - \sqrt{1 + x^2} \right)$, then

$
\begin{equation}
\begin{aligned}
y' &= \frac{1}{1+\left( x- \sqrt{1+x^2} \right)^2} \cdot \frac{d}{dx} \left( x - \sqrt{1+x^2} \right)\\
\\
y' &= \frac{1}{1 + \left( x^2 - 2x \sqrt{1+x^2} + (1 + x^2) \right)} \cdot \left( 1- \frac{2x}{2\sqrt{1+x^2}} \right)\\
\\
y' &= \frac{1- \frac{x}{\sqrt{1+x^2}}}{2 + 2x^2 - 2x \sqrt{1+x^2}}\\
\\
y' &= \frac{\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}} }{2\left( 1 + x^2 - x \sqrt{1+x^2} \right)}\\
\\
y' &= \frac{\sqrt{1+x^2}-x}{2\sqrt{1+x^2}\left( 1 + x^2 - x \sqrt{1+x^2}\right)}
\end{aligned}
\end{equation}
$