If y = tan^{-1} left[ frac{4 sin 2x}{cos 2x - | vedclass

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(C) Given $y = 	an^{-1} \left[ \frac{4 \sin 2x}{\cos 2x - 6 \sin^2 x} ight]$.Using double angle formulas,$\sin 2x = 2 \sin x \cos x$ and $\cos 2x =

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$8$

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(C) Given $y = 	an^{-1} \left[ \frac{4 \sin 2x}{\cos 2x - 6 \sin^2 x} ight]$.Using double angle formulas,$\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$.Substituting these into the expression:$y = 	an^{-1} \left[ \frac{4(2 \sin x \cos x)}{(\cos^2 x - \sin^2 x) - 6 \sin^2 x} ight] = 	an^{-1} \left[ \frac{8 \sin x \cos x}{\cos^2 x - 7 \sin^2 x} ight]$.Divide numerator and denominator by $\cos^2 x$:$y = 	an^{-1} \left[ \frac{8 	an x}{1 - 7 	an^2 x} ight]$.Let $f(x) = \frac{8 	an x}{1 - 7 	an^2 x}$. Then $y = 	an^{-1}(f(x))$.$\frac{dy}{dx} = \frac{1}{1 + (f(x))^2} \cdot f'(x)$.At $x = 0$,$f(0) = 0$,so $\frac{dy}{dx} = f'(0)$.Using the quotient rule for $f(x) = \frac{u}{v}$,$f'(x) = \frac{u'v - uv'}{v^2}$.$u = 8 	an x \implies u' = 8 \sec^2 x$.$v = 1 - 7 	an^2 x \implies v' = -14 	an x \sec^2 x$.At $x = 0$,$u(0) = 0, u'(0) = 8, v(0) = 1, v'(0) = 0$.$f'(0) = \frac{8(1) - 0(0)}{1^2} = 8$.Thus,$\frac{dy}{dx}$ at $x = 0$ is $8$.

If $y = \tan^{-1} \left[ \frac{4 \sin 2x}{\cos 2x - 6 \sin^2 x} \right]$,then $\frac{dy}{dx}$ at $x = 0$ is

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