$\frac{d}{dx} \sin^{-1} (3x-4x^3)$ is equal to :

Question

( A ) $\frac{3}{\sqrt{1-x^2}}$
( B ) $\frac{1}{\sqrt{4-x^2}}$
( C ) $\frac{3}{\sqrt{4-x^2}}$
( D ) $-\frac{1}{\sqrt{4-x^2}}$