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Find $\frac{dy}{dx}$ if $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$

The solution is given as $y'=0$ in my reference. But that doesn't seem to be a complete solution as the graph of the function is: enter image description here

My Attempt

Let $x=\sin\alpha\implies \alpha=\sin^{-1}x$ $$ y=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{1-\sin^2\alpha}=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\cos^2\alpha}\\ =\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\sin^2(\tfrac{\pi}{2}-\alpha)}=\sin^{-1}(\sin\alpha)+\sin^{-1}|\sin(\tfrac{\pi}{2}-\alpha)|\\ =n\pi+(-1)^n(\alpha)+ $$

How do I proceed further and find the derivative ?

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The derivative is: (for positive x)

$$y^{\prime}=\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{\frac{-2x}{2\sqrt{1-x^{2}}}}{\sqrt{1-1+x^{2}}}=\dfrac{1}{\sqrt{1-x^{2}}}+\dfrac{-1}{\sqrt{1-x^{2}}}=0$$

if x<0 we have : $$y^{\prime}=\dfrac{2}{\sqrt{1-x^{2}}}$$

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  • $\begingroup$ Only for $x\geq0$. $\endgroup$ – Przemysław Scherwentke Mar 31 '18 at 4:20
  • $\begingroup$ Because $\sqrt{x^2}\neq x$. $\endgroup$ – Przemysław Scherwentke Mar 31 '18 at 4:21
  • $\begingroup$ the derivative of $sin^{-1}(f(x))$ is $\dfrac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}$ $\endgroup$ – shere Mar 31 '18 at 4:22
  • $\begingroup$ @shere Your answer is wrong for $x<0$ because there's a corner and it changes derivatives at $0$ as per PrzemysławScherwentke $\endgroup$ – Andrew Li Mar 31 '18 at 4:24
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    $\begingroup$ Read the OP's request to do this with substitution instead... $\endgroup$ – Andrew Li Mar 31 '18 at 4:34
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From the end of your second line you obtain $$ \sin^{-1}(\sin\alpha)+\sin^{-1}\left(\sin\left(\frac\pi2-\alpha\right)\right)=\alpha+\frac\pi2-\alpha=\frac\pi2 $$ for $\alpha\in[0,\pi/2]$ and $$ \alpha+\alpha-\frac\pi2=2\sin^{-1}(x)-\frac\pi2 $$ for $\alpha\in[-\pi/2,0]$, which gives the same derivatives as in a standard method.

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$y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$

Let, $x=\sin\alpha\implies \alpha=\sin^{-1}x$, We have $-\pi/2\leq\alpha\leq\pi/2\implies|\cos\alpha|=\cos\alpha$ $$ \begin{align} y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(|\cos\alpha|)=\sin^{-1}(\sin\alpha)+\sin^{-1}(\cos\alpha)\\&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha)) \end{align} $$ Here, $$ \tfrac{-\pi}{2}\leq\alpha\leq\tfrac{\pi}{2}\implies\sin^{-1}(\sin\alpha)=\alpha\\ 0\leq\tfrac{\pi}{2}-\alpha\leq{\pi}\implies\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))=\begin{cases}\frac{\pi}{2}-\alpha,\text{ if }0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\ \pi-(\frac{\pi}{2}-\alpha),\text{ if }\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi\end{cases} $$ Therefore, $$ \begin{align} y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))\\ &=\begin{cases}\alpha+\frac{\pi}{2}-\alpha=\frac{\pi}{2}\quad\quad\quad\quad\;\;\text{ if }\quad\: 0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\ \alpha+\pi-\frac{\pi}{2}+\alpha=\frac{\pi}{2}+2\alpha\quad\text{ if }\quad\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi \end{cases}\\ &=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\text{ if }\quad0\leq\alpha\leq\tfrac{\pi}{2}\\ \frac{\pi}{2}+2\alpha\quad\text{ if }\quad \frac{-\pi}{2}\leq\alpha<0 \end{cases}\\ &=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\quad\quad\text{ if }\quad\quad 0\leq x\leq 1\\ \frac{\pi}{2}+2\sin^{-1}x\quad\text{ if }\quad -1\leq x<0 \end{cases} \end{align} $$ $$ \color{red}{ \frac{dy}{dx}=\begin{cases}0\quad\quad\quad\text{ if }\;\quad 0\leq x\leq 1\\ \frac{2}{\sqrt{1-x^2}}\quad\;\text{ if }\; -1\leq x<0 \end{cases}} $$

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