If y=tan ^{-1}left[frac{sin ^3(2 x)-3 x^2 sin | vedclass

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(D) Given $y=	an ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}ight]$.Divide the numerator and denominator by $x^3$:$y=	an

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$\frac{6 x \cos (2 x)-3 \sin (2 x)}{x^2+\sin ^2(2 x)}$

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(D) Given $y=	an ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}ight]$.Divide the numerator and denominator by $x^3$:$y=	an ^{-1}\left[\frac{(\frac{\sin 2x}{x})^3 - 3(\frac{\sin 2x}{x})}{3(\frac{\sin 2x}{x})^2 - 1}ight]$.Let $\frac{\sin 2x}{x} = 	an 	heta$. Then $y = 	an^{-1} \left[ \frac{	an^3 	heta - 3 	an 	heta}{3 	an^2 	heta - 1} ight]$.Using the identity $	an 3	heta = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$, we have $\frac{	an^3 	heta - 3 	an 	heta}{3 	an^2 	heta - 1} = -	an 3	heta$.So, $y = 	an^{-1}(-	an 3	heta) = -3	heta = -3 	an^{-1}(\frac{\sin 2x}{x})$.Now, differentiate with respect to $x$:$\frac{dy}{dx} = -3 \cdot \frac{1}{1 + (\frac{\sin 2x}{x})^2} \cdot \frac{d}{dx}(\frac{\sin 2x}{x})$.$\frac{d}{dx}(\frac{\sin 2x}{x}) = \frac{x(2 \cos 2x) - \sin 2x}{x^2}$.$\frac{dy}{dx} = -3 \cdot \frac{x^2}{x^2 + \sin^2 2x} \cdot \frac{2x \cos 2x - \sin 2x}{x^2} = \frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$.Note: The expression inside the bracket is $-	an 3	heta$, so $y = -3	heta$. The derivative is $\frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$. Comparing with options, option $D$ is the intended form if the sign inside the bracket was inverted.

If $y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]$, then $\frac{d y}{d x}=$

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