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Differentiate $\sin^{-1} \sqrt {1-x^2} $ wrt $\cos^{-1} $ if $x\in (-1,0)$

Let $x=\cos y$

Then $f(x)= \sin^{-1} \sqrt{1-\cos^2y}=\sin^{-1} |\sin y| =\pm y$

And $g(x)=\cos^{-1} \cos y =y$

$$\frac{f’(x)}{g’(x)} = \pm 1$$

Since $\sin y$ can be both positive or negative if $\cos y \in (-1,0)$

Which is the right answer?

V.G
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Aditya
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  • $\sin^{-1}|\sin y| \neq \pm y$ and similarly $\cos^{-1}\cos y \neq y$. For example, take $y=4\pi$, does it satisfy your equation? – V.G Aug 25 '20 at 09:51
  • Okay, he should have mentioned that $y\in (\pi/2,\pi)$ when letting $x=\cos y$. – V.G Aug 25 '20 at 10:02
  • I want to just clear this point to @Aditya in general that $\sin^{-1}\sin x \ne x$ and similarlly for $\cos $, See my answer here. – V.G Aug 25 '20 at 10:08
  • @ABCD, then the substitution doesn't make any sense because in order to perform substitution, it must be bijective mapping. OP surely would have confused that $\cos^{-1}x=y\implies x=\cos y$ which is not correct. – SarGe Aug 25 '20 at 10:09
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    @SarGe, yeah it seems so, that's why I provided a link for seeing how to deal with inverse functions. Irony is, the answer to the question which I have written there, is asked by him only! – V.G Aug 25 '20 at 10:11
  • @SarGe y can be $y\in (\pi/2, 3\pi /2)$ and $\sin^{-1} \sin y =3\pi/2-y$ doesn’t seem to work for the domain you picked – Aditya Aug 25 '20 at 10:33
  • @Aditya, my mistake. It should be $\pi-y$. – SarGe Aug 25 '20 at 10:37

2 Answers2

1

You can simply go for direct differentiation

$$\frac{d(\sin^{-1}\sqrt{1-x^2})}{d(\cos^{-1}x)}\begin{align}&=\frac{\frac{d(\sin^{-1}\sqrt{1-x^2})}{dx}}{\frac{d(\cos^{-1}x)}{dx}}\\ &=\frac{\frac{1}{\sqrt{1-(1-x^2)}}\cdot\frac{1}{2\sqrt{1-x^2}}\cdot(-2x)}{\frac{-1}{\sqrt{1-x^2}}}\\ &=\frac{x}{|x|}\\ &=-1\qquad(\because x<0\implies|x|=-x) \end{align}$$

SarGe
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1

For real $x,\sqrt{1-x^2}\ge0$

Let $\sin^{-1}\sqrt{1-x^2}=y,\implies y\ge0, x^2=\cos^2y$ and $0\le y\le\dfrac\pi2$

As $x<0,\cos y=-x$

$y=\cos^{-1}(-x)=\pi-\cos^{-1}x$

using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?

lab bhattacharjee
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