Trig differentiation Prove that there is a
constant C such that
$$ \arcsin{\frac{1-x}{1+x}} + 2\arctan (\sqrt{x})  = C $$
for all $x$ in a certain domain. What is the largest domain on which this identity is
true? What is the value of the constant $C$?
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Well, we can show that they differ by a constant if we show that $\arcsin\frac{1-x}{1+x}$ and $-2\arctan (x^{1/2}) $ have the same derivatives
I am trying to do that and i get $$ (1+x)/2(x^{1/2})$$ for arcsin and
$$ -2/(2(x^{1/2}) + 2x^{3/2}) $$ for arctan
that are not equal... what have I done wrong

\begin{align}
& \frac d{dx} \arcsin\frac{1-x}{1+x} = \frac 1 {\sqrt{1-\left( \frac{1-x}{1+x} \right)^2}} \cdot \frac d {dx} \frac{1-x}{1+x} \\[10pt]
= {} & \frac 1 {\sqrt{1-\left( \frac{1-x}{1+x} \right)^2}} \cdot \frac{-2}{(1+x)^2} = \frac{1+x}{\sqrt{(1+x)^2- (1-x)^2}} \cdot\frac{-2}{(1+x)^2} \\[10pt]
= {} & \frac{-1}{(1+x)\sqrt{x}} \\[10pt]
& \text{and} \\[10pt]
& \frac d{dx} \arctan\sqrt x = \frac{1}{1+x}\cdot\frac d {dx} \sqrt x = \frac 1 {1+x} \cdot\frac 1 {2\sqrt x}
\end{align}
This is valid on an interval where $1+x$ is positive (otherwise where we had $\sqrt{(1+x)^2}$, it would have been $|1+x|$, etc.).  And we must also have $x\ge 0$.
One should also figure out when $\dfrac{1-x}{1+x}$ is in the domain of the arcsine function.
