# derivative of Inverse Trigonometric problems

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

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?

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

Always keep in mind the principal values of

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

$$-\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 $$y$$

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