If $x=\int\limits_0^y\frac{dt}{\sqrt{1+9t^2}}$ and $\frac{d^2y}{dx^2}$; then ‘a’ is equal to:

3 6 9 1

9

$\frac{dx}{dy}=\frac{1}{\sqrt{1+9y^2}}⇒\frac{dy}{dx}=\sqrt{1+9y^2}⇒\frac{d^2y}{dx^2}=\frac{9.2.y}{2\sqrt{1+9y^2}}.\frac{dy}{dx}$ $⇒\frac{d^2y}{dx^2}=\frac{9y}{\sqrt{1+9y^2}}.\frac{\sqrt{1+9y^2}}{1}=9y⇒a=9$