Question

Let $y\left(x\right)=cos\left(3co{s}^{-1}x\right),x\in [-1,1],x\ne \pm \frac{\sqrt{3}}{2}$, Then $\frac{1}{y\left(x\right)}\left\{\right(x^2-1)\frac{d^{2}y\left(x\right)}{d{x}^2}+x\frac{dy\left(x\right)}{dx}\}$ equal to?

Answer 1

$y=\cos \left(3{\cos }^{-1}x\right)$

$=\cos \left({\cos }^{-1}\right(4x^3-3x)$

$y=4x^3-3x$

$\frac{dy}{dx}=12x^2-3$

$\frac{d^{2}y}{d{x}^2}=24x$

Put these values in given expression

$=\frac{1}{4x^3-3x}\left[\right(x^2-1)24x+x(12x^2-3\left)\right]$

$=\frac{1}{4x^3-3x}[24x^3-24x+12x^3-3x]$

$=\frac{1}{4x^3-3x}[36x^3-27x]$

$=9\frac{[4x^3-3x]}{(4x^3-3x)}$

$=9$