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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 ?

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Note that

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

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

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  • $\begingroup$ could u comment on my attempt ?. why am i getting two case ? $\endgroup$ – ss1729 Mar 23 '18 at 17:09
  • $\begingroup$ @ss1729 so you are looking for a direct proof? $\endgroup$ – user Mar 23 '18 at 17:10
  • $\begingroup$ @ss1729 the final result seems to be correct, but why use the $tan^{-1}$ for the derivation? $\endgroup$ – user Mar 23 '18 at 17:15
  • $\begingroup$ Why do toy thunk that you should be getting one case? $\endgroup$ – Michael McGovern Mar 23 '18 at 17:16
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    $\begingroup$ @ss1729 yes by chain rule I've derived exactly your result $\endgroup$ – user Mar 23 '18 at 17:36
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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 $$\sin^2(y)+\cos^2(y)=1$$

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  • $\begingroup$ hw do u proceed it without substituting for x ? $\endgroup$ – ss1729 Mar 23 '18 at 17:27
  • $\begingroup$ differentiate both sides and note that $(\sin(y))'=\cos(y)y'$ $\endgroup$ – Satyendra Mar 23 '18 at 17:29
  • $\begingroup$ what about RHS ?. $\cos yy'=\frac{(1+x^2).2-2x.2x}{(1+x^2)^2}=\frac{2+4x^2+2x^4-4x^2}{(1+x^2)^2}=\frac{2(1+x^4)}{(1+x^2)^2}$ Now what ? $\endgroup$ – ss1729 Mar 23 '18 at 17:33
  • $\begingroup$ you have the value of $\sin(y)$ use $cos^2(y)+sin^2(y)=1$ to get rid of the $cos(y)$ $\endgroup$ – Satyendra Mar 23 '18 at 17:35
  • $\begingroup$ This answer is good when you dont know the derivative of arcsin function .... $\endgroup$ – Satyendra Mar 23 '18 at 19:33

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