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y=cos−1cos(log22lnesin−1sinx)\large y = \cos^{-1}\cos(\log_{2}2^{\ln e^{\sin^{-1}\sin x}})y=cos−1cos(log22lnesin−1sinx)
For yyy as defined above, find dydx\dfrac{dy}{dx}dxdy at x=π4x = \dfrac{\pi}{4}x=4π.
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