Question

Differentiate ${\sin }^{-1}(3x-4x^3)$ with respect to $x$, when
$\left(i\right)\frac{1}{2}<x<1,\left(ii\right)-1<x<\frac{-1}{2}$

Answer 1

Let $y={\sin }^{-1}(3x-4x^3)$
Putting $x=\sin \theta$, we get
$y={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )={\sin }^{-1}\left(\sin 3\theta \right)$

$\left(i\right)$ When $\frac{1}{2}<x<1$, then
$x=\sin \theta \Rightarrow \frac{1}{2}<\sin \theta <1$
$\Rightarrow \frac{\pi }{6}<\theta <\frac{\pi }{2}$
$\Rightarrow \frac{\pi }{2}<3\theta <\frac{3\pi }{2}$
$\Rightarrow -\frac{\pi }{2}>-3\theta >-\frac{3\pi }{2}$
$\Rightarrow -\frac{\pi }{2}<\pi -3\theta <\frac{\pi }{2}$

$y={\sin }^{-1}\left(\sin 3\theta \right)$
$\Rightarrow y={\sin }^{-1}\left(\sin \right(\pi -3\theta \left)\right)$
$\Rightarrow y=\pi -3\theta$
$\Rightarrow y=\pi -3{\sin }^{-1}x$
$\frac{dy}{dx}=0-\frac{3}{\sqrt{1-x^2}}=\frac{-3}{\sqrt{1-x^2}}$

$\left(ii\right)$ When $-1<x<\frac{-1}{2}$
$x=\sin \theta \Rightarrow -1<\sin \theta <-\frac{1}{2}$
$\Rightarrow -\frac{\pi }{2}<\theta <-\frac{\pi }{6}$
$\Rightarrow -\frac{3\pi }{2}<3\theta <-\frac{\pi }{2}$
$\Rightarrow \frac{\pi }{2}<-3\theta <\frac{3\pi }{2}$
$\Rightarrow -\frac{\pi }{2}<-\pi -3\theta <\frac{\pi }{2}$
$y={\sin }^{-1}\left(\sin 3\theta \right)$
$y={\sin }^{-1}\left(\sin \right(-\pi -3\theta \left)\right)$
$y=-\pi -3\theta$
$y=-\pi -3{\sin }^{-1}x$
$\frac{dy}{dx}=-0-\frac{3}{\sqrt{1-x^2}}=-\frac{3}{\sqrt{1-x^2}}$