Im trying to find the derivative of $\arctan(x-\sqrt{x^2+1})$ here are my steps if someone could point out where I went wrong.
$$\begin{align}
\frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x}
\\[1ex]
& =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\sqrt{x^2+1})^2}}
\end{align}$$
Everything after this turns into a huge mess I don't know how to simplify. Is there a trick I missed or something I don't see?
Best Answer
The expression simplifies very nicely.
Expand the bottom. We get $2\sqrt{x^2+1}(\sqrt{x^2+1}-x)$. Note that the top is $\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}}$. There is cancellation, and we end up with $\frac{1}{2(x^2+1)}$.