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I am unable to understand why the derivative of : $$y = \sin^{-1}(2x\sqrt {1-x^2}), {-1\over\sqrt{2}}<x<{1\over\sqrt{2}} $$

is: $$ 2\over\sqrt {1-x^2} $$

and why it cannot be / is not : $$ -2\over\sqrt {1-x^2} $$

well, if we take $x$ as $\cos\theta$ and proceed...

$$ y = \sin^{-1}(2 \cos\theta\sqrt {1-\cos^2\theta}) $$

$$ \Rightarrow y =\sin^{-1}(2\sin\theta \cos\theta) $$

$$ \Rightarrow y = \sin^{-1}(\sin2\theta)$$

$$ \Rightarrow y = 2\theta $$

$$ \Rightarrow y = 2\cos^{-1}x $$

the derivative comes out to be: $$ -2\over\sqrt {1-x^2} $$

Now, at first, I thought that it's due to the domain inequality ${-1\over\sqrt{2}}<x<{1\over\sqrt{2}}$ (if we take the $\cos^{-1}$ of this inequality, we get an absurd solution. $({3\pi \over 4} < \cos^{-1}x < {\pi \over 4})$.

Certainly, ${3\pi \over 4} < {\pi \over 4}$ is False. So I thought this was the reason why we cant take $x$ as $\cos\theta$. ( we take $x$ as $\sin\theta$ and get the right answer $2\over\sqrt {1-x^2}$

Then, the next question was to find the derivative of:

$$ y = \sec^{-1}{1 \over (2x^2 - 1)} , 0<x<{1\over\sqrt{2}} $$

Here taking $x$ as $\cos\theta$ gives the absurdity again, BUT gives the answer too... So, after all, what is the correct way to solve these problems?

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  • $\begingroup$ math.stackexchange.com/questions/672575/… $\endgroup$ – lab bhattacharjee Jun 4 at 4:07
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    $\begingroup$ When $x=\cos\theta$, the continuous branch of $y=\sin^{-1}\sin(2\theta)$ is not $2\theta$, but $\pi-2\theta$. You can see this by trying $x=0$, so $\theta=\pi/2$ and $y=\sin^{-1}(\sin\pi)=0$. $\endgroup$ – user10354138 Jun 4 at 4:09
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    $\begingroup$ $\sqrt {1-\cos^2 \theta} = \sqrt {\sin^2 \theta} = |\sin \theta|.$ In this case $-\frac {\pi} {4} < \theta < \frac {\pi} {4}.$ In this range of $\theta,$ $\sin \theta \not\geq 0.$ Therefore $|\sin \theta| \neq \sin \theta.$ $\endgroup$ – Dbchatto67 Jun 4 at 4:10
  • $\begingroup$ Please share the details of the last absurdity $\endgroup$ – lab bhattacharjee Jun 4 at 4:20
  • $\begingroup$ @labbhattacharjee to solve the second problem, when we put x as $cos\theta$ and then take the cos inverse, we get pi < pi/4 .... $\endgroup$ – J Shelly Jun 4 at 4:36
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Always keep in mind the principal values of

https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions

$-\dfrac1{\sqrt2} <x<\dfrac1{\sqrt2}$

actually implies $\dfrac{3\pi}4>\arccos x=y>\dfrac\pi4$ as $\arccos(x)$ is decreasing in $[-\dfrac\pi2,\dfrac\pi2]$

$\implies?>2y>\dfrac\pi2$

But $-\dfrac\pi2\le u=\arcsin(\sin2y)\le\dfrac\pi2$

So,$u=\pi-2y$ for the above range of $y$

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  • $\begingroup$ I did not understand this... $\endgroup$ – J Shelly Jun 4 at 4:34
  • $\begingroup$ @JShelly, please pinpoint the confusion $\endgroup$ – lab bhattacharjee Jun 4 at 4:46
  • $\begingroup$ the 2nd line "actually implies...." $\endgroup$ – J Shelly Jun 4 at 5:04
  • $\begingroup$ @JShelly, If $x<\dfrac1{\sqrt2},\cos^{-1}x=\arccos(x)>\dfrac\pi4$ right? $\endgroup$ – lab bhattacharjee Jun 4 at 5:14
  • $\begingroup$ yes, you're right $\endgroup$ – J Shelly Jun 4 at 6:51

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