derivative of arctan(u)
Sebastian Wright 
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?
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$\begingroup$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 $\frac{1}{2(x^2+1)}$.
$\endgroup$ 0 $\begingroup$It seems about right. $$\begin{align} \frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x} \\[1ex] & =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\sqrt{x^2+1})^2}} \\[1ex] & =\; \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}\;(1+x^2-2x\sqrt{x^2+1}+x^2+1)} \\[1ex] & =\; \frac{\sqrt{x^2+1}-x}{2\sqrt{x^2+1}\;(1+x^2-x\sqrt{x^2+1})} \\[1ex] & =\; \frac{(\sqrt{x^2+1}-x)\sqrt{x^2+1}\;(1+x^2+x\sqrt{x^2+1})}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))} \\[1ex] & =\; \frac{(x^2+1-x\sqrt{x^2+1})\;(1+x^2+x\sqrt{x^2+1})}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))} \\[1ex] & =\; \frac{((1+x^2)^2-x^2(x^2+1))}{2(x^2+1)\;((1+x^2)^2-x^2(x^2+1))} \\[2ex] & =\; \frac{1}{2(x^2+1)} \end{align}$$
[edit]okay, turns out it is a neat derivative[/edit]
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