If y=tan ^{-1}left(frac{4 sin 2 x}{cos 2 x-6 | vedclass

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(C) Given $y=	an ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}ight)$.Using the identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x

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$8$

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(C) Given $y=	an ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}ight)$.Using the identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$:$y = 	an^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - \sin^2 x - 6 \sin^2 x} ight) = 	an^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - 7 \sin^2 x} ight)$.Divide numerator and denominator by $\cos^2 x$:$y = 	an^{-1} \left( \frac{8 	an x}{1 - 7 	an^2 x} ight)$.We can write $8 	an x = 7 	an x + 	an x$:$y = 	an^{-1} \left( \frac{7 	an x + 	an x}{1 - (7 	an x)(	an x)} ight)$.Using the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$:$y = 	an^{-1}(7 	an x) + 	an^{-1}(	an x) = 	an^{-1}(7 	an x) + x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{1 + (7 	an x)^2} \cdot 7 \sec^2 x + 1 = \frac{7 \sec^2 x}{1 + 49 	an^2 x} + 1$.At $x=0$,$	an 0 = 0$ and $\sec 0 = 1$:$\left(\frac{dy}{dx}ight)_{x=0} = \frac{7(1)^2}{1 + 49(0)^2} + 1 = \frac{7}{1} + 1 = 8$.

If $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$,then $\left(\frac{d y}{d x}\right)$ at $x=0$ is

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