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If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to

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If $y = \sec(\tan^{-1} x)$,then $\frac{dy}{dx}$ at $x = 1$ is

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