If $y=sin^{-1}x,$ then $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
$y=sin^{-1}x⇒\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$
so $\frac{d^2y}{dx^2}=\frac{-1}{2\sqrt{1-x^2}}\frac{(-2x)}{(1-x^2)}=\frac{x}{\sqrt{1-x^2}(1-x^2)}$
so $(1-x^2)\frac{d^2y}{dx^2}=x\frac{dy}{dx}$
$⇒(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0$