Problem
Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, where $y = \sin^{-1}\biggl(\dfrac{2x}{1+x^2}\biggr)$
Given solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{1+x^2}$
My approach
I got the solution as $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2}{1+x^2}$ (negative of the given solution).
I took $t = \dfrac{2x}{1+x^2}$, so $\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{2(1-x^2)}{(1+x^2)^2}$.
Then I took $s = \sin^{-1}t$, so $\dfrac{\mathrm{d}s}{\mathrm{d}t} = \dfrac{1}{(1-t^2)^{1/2}} = \dfrac{(x^2+1)}{(x-1)(x+1)}$ so $$ \dfrac{\mathrm{d}s}{\mathrm{d}t} = 2\frac{1-x^2}{(1+x^2)^2} \cdot \frac{x^2+1}{(x-1)(x+1)} = 2\frac{1-x}{(x-1)(1+x^2)} = -\frac{2}{1+x^2}. $$
I googled solutions for this, most of them proceeded with substituting $x = \tan\theta$. I want to know why I'm getting negative of the actual answer, OR is both of these are the solutions for this?
My guess is that both positive and negatives are the solutions for this. In my approach I factorised the term $(1+x^2)^2 - (2x)^2$ under square root, which you would get after substituting $t$ with $$ \frac{2x}{1+x^2} $$ in $\frac{\mathrm{d}s}{\mathrm{d}t}$, as $(x-1)^2 \cdot (x+1)^2$, therefore I ended up with factors $(1-x)$ and $(x-1)$ in numerator and denominator of the solution, respectively. If I instead factorised it as $(1-x)^2 \cdot (x+1)^2$, then I would have got the solution exact to the given one.
You are right. Both signs are correct.
The function has max/min at $x=\pm 1 $.
In the interval $|x|<1$ the slope is positive.
Outside this interval it is negative.
This often happens for inverse functions.
Curious fact
In the first quadrant slope changes $(1\to 0 \to -1)$ at maximum point $ (1,\dfrac{\pi}{2}).$