Differentiate sin⁡−1(2x1+x2)\sin^{-1} \left( \frac{2x}{1 + x^2} \right)sin−1(1+x22x) with respect to xxx.

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Differentiate $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ $sin$ $-1$ $($ $1$ $+$ $x$ $2$ $2$ $x$ $$ $)$ with respect to $x$ $x$ .

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Detailed Solution

Differentiate $\displaystyle \sin^{-1}\!\left(\frac{2x}{1+x^{2}}\right)$ $sin$ $-1$ $($ $1$ $+$ $x$ $2$ $2$ $x$ $$ $)$

Let $u=\dfrac{2x}{1+x^2}$ $u$ $=$ $1$ $+$ $x$ $2$ $2$ $x$ $$ .

u'=\frac{2(1-x^2)}{(1+x^2)^2},\quad 1-u^2=\frac{(1-x^2)^2}{(1+x^2)^2}

$u$ $'$ $=$ $($ $1$ $+$ $x$ $2$ $)$ $2$ $2$ $($ $1$ $-$ $x$ $2$ $)$ $$ $,$ $1$ $-$ $u$ $2$ $=$ $($ $1$ $+$ $x$ $2$ $)$ $2$ $($ $1$ $-$ $x$ $2$ $)$ $2$ $$ $\frac{d}{dx}\big[\sin^{-1}(u)\big] =\frac{u'}{\sqrt{1-u^2}} =\frac{\tfrac{2(1-x^2)}{(1+x^2)^2}}{\tfrac{|1-x^2|}{1+x^2}} =\frac{2}{1+x^2}\cdot \frac{1-x^2}{|1-x^2|}$ $dxd$ $$ $[$ $sin$ $-1$ $($ $u$ $)$ $]$ $=$ $1$ $-$ $u$ $2$ $$ $u$ $'$ $$ $=$ $1$ $+$ $x$ $2$ $∣1$ $-$ $x$ $2$ $∣$ $$ $($ $1$ $+$ $x$ $2$ $)$ $2$ $2$ $($ $1$ $-$ $x$ $2$ $)$ $$ $$ $=$ $1$ $+$ $x$ $2$ $2$ $$ $\cdot$ $∣1$ $-$ $x$ $2$ $∣1$ $-$ $x$ $2$ $$

So,

$\boxed{\,y'=\frac{2}{1+x^2}\, \text{for }|x|<1;\quad y'=-\frac{2}{1+x^2}\, \text{for }|x|>1;\quad \text{undefined at }x=\pm1\,}$ $y$ $'$ $=$ $1$ $+$ $x$ $2$ $2$ $$ $for$ $∣$ $x$ $∣$ $<$ $1$ $;$ $y$ $'$ $=$ $-$ $1$ $+$ $x$ $2$ $2$ $$ $for$ $∣$ $x$ $∣$ $>$ $1$ $;$ $undefined at$ $x$ $=$ $\pm1$ $$

(Using the identity $\sin^{-1}\!\big(\frac{2x}{1+x^2}\big)=2\tan^{-1}x$ $sin$ $-1$ $($ $1$ $+$ $x$ $2$ $2$ $x$ $$ $)$ $=$ $2$ $tan$ $-1$ $x$ on $|x|<1$ $∣$ $x$ $∣$ $<$ $1$ gives $y'=\frac{2}{1+x^2}$ $y$ $'$ $=$ $1$ $+$ $x$ $2$ $2$ $$ immediately.)