$\begin{matrix} Find the derivative of the function given by f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) and hence find f^{'} (1) . \end{matrix}$

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Solution

$f (n) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$ and hence
find $f^{- 1} (1)$
$f^{1} (n) = (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) + 2 x (1 + x) (1 + x^{4}) (1 + x^{8})$
$+ (1 + x) (1 + x^{2}) 4 x^{3} (1 + x^{8}) + 8 x^{7} (1 + x) (1 + x^{2}) (1 + x^{4})$
$f^{1} (1) = 2.2.2 + 2.2.2.2 + 2.2.4.2 + 8.2.2.2.$
$= 2.2.2 [1 + 2 + 4 + 8]$
$= 8 [5] = 120$ Ans

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