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?
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 $y$