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

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

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}ight) \ & =	an ^{-1}\left[\frac{4(2 \sin x \cos x)}{\left(\cos ^2 x-\sin ^2 xight)-6 \sin ^2 x}ight] \ & =	an ^{-1}\left(\frac{8 \sin x \cos x}{\cos ^2 x-7 \sin ^2 x}ight)\end{aligned}$ $\begin{aligned} & =	an ^{-1}\left(\frac{8 	an x}{1-7 	an ^2 x}ight) \ & =	an ^{-1}\left(\frac{7 	an x+	an x}{1-7 	an x \cdot 	an x}ight) \ & =	an ^{-1}(7 	an x)+	an ^{-1}(	an x) \ 	herefore \quad & y=	an ^{-1}(7 	an x)+x \ 	herefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+(7 	an x)^2} \cdot 7 \sec ^2 x+1\end{aligned}$ $\begin{gathered}=\frac{7 \sec ^2 x}{1+49 	an ^2 x}+1 \ 	herefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=0}= \ =\frac{7 \sec ^2 0}{1+49 	an ^2 0}+1 \ =\frac{7}{1+0}+1 \ =8\end{gathered}$

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

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