If $y=\tan ^{-1}\left(\frac{3 \cos x-4 \sin x}{4 \cos x+3 \sin x}\right)+2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$, then $\frac{d y}{d x}$ at $x=\frac{\sqrt{3}}{2}$ is equal to:
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