Find derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$

Question

Find $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$

The solution is given as $\frac{2}{1+x^2}$. But is it a complete solution ?

My Attempt

$$ 2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2},\quad |x|\leq 1\\ \pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\; x>0\\ -\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\;x>0\\ \end{cases}\\ \sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x,\quad |x|\leq 1\\ \pi-2\tan^{-1}x,\quad |x|>1 \;\&\; x>0\\ -\pi-2\tan^{-1}x,\quad |x|>1 \;\&\;x>0\\ \end{cases}\\ $$ Thus, $$ \frac{dy}{dx}=\frac{d}{dx}\bigg[\sin^{-1}\frac{2x}{1+x^2}\bigg]=\begin{cases} \frac{d}{dx}[2\tan^{-1}x]=\frac{2}{1+x^2},\quad |x|\leq 1\\ \frac{d}{dx}[\pm\pi-2\tan^{-1}x]=\frac{-2}{1+x^2},|x|>1 \end{cases} $$ Is it correct ?

Answer 1

Note that

• $(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$

then apply chain rule $f(g(x))'=f'(g(x))g'(x)=\frac{\frac{2(1+x^2)-4x^2}{(1+x^2)^2}}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}=\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$

Answer 2

Hint Your answer seems correct to me even I don't really understand the need of tan function

$$y=\sin^{-1}\frac{2x}{1+x^2} \implies \sin y=\frac{2x}{1+x^2}$$ Differentiate $$y'\cos(y)=\frac {-2(x^2-1)}{(x^2+1)^2}$$ use the fact that $$\sin^2(y)+\cos^2(y)=1$$