# Differentiate $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

Find $\dfrac{\mathrm dy}{\mathrm dx}$ if $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

I can solve it as follows: \begin{align} y'&=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\Big(2x\sqrt{1-x^2}\Big)\\&=\frac{1}{\sqrt{1-4x^2+4x^4}}\bigg(2x\frac{-2x}{2\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg) \\ &=\frac{1}{\sqrt{(1-2x^2)^2}}\bigg(\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg)\\ &=\frac{1}{|1-2x^2|}\frac{-2x^2+2-2x^2}{\sqrt{1-x^2}}\\ &=\frac{2(1-2x^2)}{|1-2x^2|\sqrt{1-x^2}}\\ \end{align} As $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies|x|<\frac{1}{\sqrt{2}}\implies 0<x^2<\frac{1}{2}\implies 0<2x^2<1\\\implies-1<-2x^2<0\implies 0<1-2x^2<1$

Thus, $|1-2x^2|=1-2x^2$ $$y'=\frac{2(1-2x^2)}{(1-2x^2)\sqrt{1-x^2}}=\frac{2}{\sqrt{1-x^2}}$$

My Attempt

But if I try to solve it by substituting $x=\sin\alpha\implies\alpha=\sin^{-1}x$ $$y=\sin^{-1}\bigg(2\sin\alpha\sqrt{\cos^2\alpha}\bigg)=\sin^{-1}\bigg(2\sin\alpha|\cos\alpha|\bigg)$$

Here, $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies \frac{-\pi}{4}<\sin^{-1}x=\alpha<\frac{\pi}{4}\implies \cos\alpha>0\implies|\cos\alpha|=\cos\alpha$

\begin{align} y&=\sin^{-1}\bigg(2\sin\alpha\cos\alpha\bigg)=\sin^{-1}\bigg(\sin2\alpha\bigg)\\&\implies\sin y=\sin2\alpha=\sin\Big(2\sin^{-1}x\Big)\\ &\implies y=n\pi+(-1)^n(2\sin^{-1}x)=\begin{cases}n\pi+2\sin^{-1}x,\quad \text{n even}\\ n\pi-2\sin^{-1}x,\quad \text{n odd} \end{cases} \end{align}

Thus,

$$y'=\begin{cases}\frac{d}{dx}\Big(n\pi+2\sin^{-1}x\Big)=\frac{2}{\sqrt{1-x^2}},\quad \text{n even}\\ \frac{d}{dx}\Big(n\pi-2\sin^{-1}x\Big)=\frac{-2}{\sqrt{1-x^2}},\quad \text{n odd} \end{cases}$$

How do I eliminate the case $y'=\frac{-2}{\sqrt{1-x^2}}$ ?

• why not by chain rule? – gimusi Mar 24 '18 at 10:00

At one point you have $y=\sin^{-1}(\sin 2\alpha)$. It means that $-\pi/2\le y\le \pi/2$, so your only allowed solution is $n=0$.

• ok. i think its bcz $-\pi/2\leq y\leq\pi/2$ and $\frac{-\pi}{4}<\sin^{-1}x<\frac{\pi}{4}\implies\frac{-\pi}{2}<2\sin^{-1}x<\frac{\pi}{2}$ Thus $n=0$ is the only solution. $y'=\frac{d}{dx}0\pi+2\sin^{-1}x=\frac{2}{\sqrt{1-x^2}}$ right ? – ss1729 Mar 24 '18 at 10:24
• Yes. You also don't need to differentiate $n\pi=0$ – Andrei Mar 24 '18 at 10:32

The idea is very good. However, the notation $\sin^{-1}t$ usually doesn't mean “the set of all angles $\varphi$ such that $\sin\varphi=t$”, but rather

$\sin^{-1}t$ denotes the unique angle $\phi\in[-\pi/2,\pi/2]$ such that $\sin\varphi=t$.

The fact that $\sin^{-1}$ is not the inverse function of the sine is the reason why many people, including myself, prefer to use $\arcsin t$ instead of the logically wrong $\sin^{-1}t$. It is the inverse function of a restriction of the sine function.

This way, for $\varphi\in[-\pi/2,\pi/2]$ and $t\in[-1,1]$ it holds that $$\sin(\sin^{-1}t)=t \qquad \sin^{-1}(\sin\varphi)=\varphi$$

If $\alpha=\sin^{-1}x$, then indeed $-\pi/4<\alpha<\pi/4$, so $\cos\alpha>0$. Thus $$\sqrt{1-x^2}=\sqrt{1-\sin^2\alpha}=\cos\alpha$$ Then $2x\sqrt{1-x^2}=2\sin\alpha\cos\alpha=\sin2\alpha$ and $-\pi/2<2\alpha<\pi/2$.

Therefore $\sin^{-1}(\sin2\alpha)=2\alpha=2\sin^{-1}x$ and so $$y'=\frac{2}{\sqrt{1-x^2}}$$

• So is it like $y=\sin^{-1}(\sin2\alpha)=2\alpha$ as $\frac{-\pi}{2}<2\alpha<\frac{\pi}{2}$. Else i need to do $\sin y=\sin2\alpha\implies y=n\pi+(-1)^n2\alpha$ and solve for each case right ? – ss1729 Mar 24 '18 at 11:32
• @ss1729 You aren't “solving the equation $\sin y=t$”, but rather computing the (single valued) function $\sin^{-1}t$. There's no $n\pi$ here. – egreg Mar 24 '18 at 11:36
• i mean what if there is no restriction on $2\alpha$, ie. not the pricipal value branch. Then we need to solve as $\sin y=\sin 2\alpha\implies y=n\pi+(-1)^n(2\alpha)$, then there will be two solutions for $y'$, i.e. $y'=\frac{\pm2}{\sqrt{1-x^2}}$ right ? – ss1729 Mar 24 '18 at 11:40
• @ss1729 No. If $\gamma$ is any angle, $\sin^{-1}(\sin\gamma)$ is the unique angle of the form $n\pi+(-1)^n\gamma$ that's inside $[-\pi/2,\pi/2]$. So you either have $+$ or $-$, but not both. – egreg Mar 24 '18 at 11:43
• i think i got confused a bit. i mean if there is no restriction on $2\alpha$ it will become $\sin y=\sin 2\alpha\implies y=n\pi+(-1)^n(2\alpha)$ thus, $y'=\frac{\pm 2}{\sqrt{1-x^2}}$ right ? – ss1729 Mar 24 '18 at 11:59