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I was required to find $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$

This is my solution.

enter image description here

Above when I put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ then I get the correct answer but when I put $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$ then I get something else.

Now my question is that which one is correct and why? How you will come to know that we should put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ and not $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$

Kindly help me.

Singh
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  • Well, $(x^2-1)^2=(1-x^2)^2$, so it doesn't matter at all. To add to this: $\frac{d}{dx}(x^2-1)^2=2(x^2-1)\cdot(2x)$, and $\frac{d}{dx}(1-x^2)^2=2(1-x^2)\cdot(-2x)=2(x^2-1)\cdot(2x)=\frac{d}{dx} (x^2-1)^2$, so I assume you may have forgotten a minus sign somewhere. – Scounged Jul 20 '15 at 16:32

2 Answers2

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HINT:

For real $a,$ $$\sqrt{a^2}=|a|$$

$=+a$ if $a\ge0$ and $=-a$ if $a<0$

Things will be clearer with the following :

Using my answer here: showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ and $\arctan y=\arcsin\dfrac y{\sqrt{y^2+1}}$

$$2\arctan x=\begin{cases} \arcsin\frac{2x}{1+x^2} &\mbox{if } x^2<1\\ \pi+\arcsin\frac{2x}{1+x^2} & \mbox{if } x^2>1\end{cases} $$

Community
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lab bhattacharjee
  • 274,582
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Hint:

Given $$ y = \sin^{-1}\left( \frac{2x}{1+x^2}\right), $$

then $$ \sin(y) = \frac{2x}{1+x^2}, $$

so $$ \cos(y) y' = \left( \frac{2x}{1+x^2} \right)'. $$

Now $$ \cos(y) = \sqrt{ 1 - \sin^2(y) } = \sqrt{ 1 - \left( \frac{2x}{1+x^2} \right)^2 }, $$

so you get $$ y' = \frac{ \displaystyle \left( \frac{2x}{1+x^2} \right)' } { \displaystyle \sqrt{ 1 - \left( \frac{2x}{1+x^2} \right)^2 }}. $$

And $$ \left( \frac{2x}{1+x^2} \right)' = 2 \frac{1 - x^2}{(1+x^2)^2}. $$

So final is $$ y' = 2 \frac{ 1 - x^2 }{ \displaystyle (1+x^2) \sqrt{ (1+x^2)^2 - 4x^2 }} = 2 \frac{ 1 - x^2 }{ \displaystyle (1+x^2) \sqrt{ (1-x^2)^2 }}. $$

Thus $$ y' = \left\{ \begin{array}{rcl} |x| < 1 : \displaystyle \frac{2}{1+x^2}\\\\ |x| > 1 : \displaystyle \frac{-2}{1+x^2} \end{array} \right. $$

johannesvalks
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