If y = |cos 4x| + |sin 4x| + |tan 4x|,then fi | vedclass

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(C) Given $y = |\cos 4x| + |\sin 4x| + |	an 4x|$.At $x = \frac{\pi}{6}$,$4x = \frac{4\pi}{6} = \frac{2\pi}{3}$.In the neighborhood of $x = \frac{\pi}

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$2\sqrt{3} - 18$

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(C) Given $y = |\cos 4x| + |\sin 4x| + |	an 4x|$.At $x = \frac{\pi}{6}$,$4x = \frac{4\pi}{6} = \frac{2\pi}{3}$.In the neighborhood of $x = \frac{\pi}{6}$,$\cos 4x  0$,and $	an 4x Therefore,$y = -\cos 4x + \sin 4x - 	an 4x$.Differentiating with respect to $x$:$\frac{dy}{dx} = -(-\sin 4x \cdot 4) + (\cos 4x \cdot 4) - (\sec^2 4x \cdot 4)$$\frac{dy}{dx} = 4\sin 4x + 4\cos 4x - 4\sec^2 4x$.At $x = \frac{\pi}{6}$,$4x = \frac{2\pi}{3}$.$\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$,$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$,$\sec(\frac{2\pi}{3}) = -2$.$\frac{dy}{dx} = 4(\frac{\sqrt{3}}{2}) + 4(-\frac{1}{2}) - 4(-2)^2$$\frac{dy}{dx} = 2\sqrt{3} - 2 - 16 = 2\sqrt{3} - 18$.Note: The provided options seem to contain a typo. Based on the calculation,the correct value is $2\sqrt{3} - 18$.

If $y = |\cos 4x| + |\sin 4x| + |\tan 4x|$,then find $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$.

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