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
 A: Let $\arctan\sqrt x=y\implies x=\tan^2y$ 
and as $\sqrt x>0, 0\le y\le\dfrac\pi2$ (See definition of principal values of $\arcsin,\arctan$)
Now $\dfrac{1-x}{1+x}=\cos2y=\sin\left(\dfrac\pi2-2y\right)$
$\implies\arcsin\dfrac{1-x}{1+x}=\dfrac\pi2-2y$  
as $0\le2y\le\pi\iff0\ge2y\ge-\pi\iff \dfrac\pi2\ge\dfrac\pi2-2y\ge-\dfrac\pi2$ which is exactly the range of principal values of $\arcsin$
A: \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.
A: Consider $\theta=\arcsin\dfrac{1-x}{1+x}$.  This is an angle for which
$$
\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1-x}{1+x}.
$$
Therefore
$$
\text{adjacent} = \sqrt{(1+x)^2-(1-x)^2} = 2\sqrt{x}.
$$
Let $\eta$ be the complementary angle $\frac\pi2-\theta$.  Then
$$
\tan\frac\eta2=\frac{\sin\eta}{1+\cos\eta} = \frac{\cos\theta}{1+\sin\theta}= \frac{2\sqrt x/(1+x)}{1+ (1-x)/(1+x)} = \frac{2\sqrt x}{2} = \sqrt x,
$$
so
$$
\frac\pi2-\theta = \eta = 2\arctan\sqrt x.
$$
This is certainly valid for $0<\theta<\pi/2$, so figure out which values of $x$ that corresponds to and also think about which other quadrants it might be valid in.
A: Let $x=\tan^2t$, and use $1+\tan^2t=\dfrac1{\cos^2t},$ $\tan t=\dfrac{\sin t}{\cos t},$ and $\cos2x=\cos^2x-\sin^2x$. This 
is justifiable by the fact that both $\sqrt x$ and $\tan t$ can take any positive real value, for $0\le t\le\dfrac\pi2$ .
