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 $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$\frac{d}{dx}\left(\frac{x^{16}-1}{x-1}\right) = \frac{(16x^{15})(x-1) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.Comparing this with the given expression $\frac{15x^p - 16x^q + 1}{(x-1)^2}$, we get $p = 16$ and $q = 15$.Thus, $(p, q) = (16, 15)$.

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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