If frac{d}{d x}left[(x+1)left(x^2+1right)left | vedclass

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(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.Multiply and divide by $(x-1)$:$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1

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$(16, 15)$

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(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.Multiply and divide by $(x-1)$:$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:$\frac{d}{dx} \left[ \frac{x^{16}-1}{x-1} \right] = \frac{(x-1)(16x^{15}) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.Comparing this with the given expression $\left(15 x^p-16 x^q+1\right)(x-1)^{-2}$,we get $p=16$ and $q=15$.Therefore,$(p, q) = (16, 15)$.

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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