Let y(x) = (1 + x)(1 + x ^ 2 )(1 + x ^ 4 )(1 + x ^ 8 )(1 + x ^ 16 )...

Given,

Function $y\left(x\right) = \left(1 + x\right) \left(1 + {x}^{2}\right) \left(1 + {x}^{4}\right) \left(1 + {x}^{8}\right)\left(1 + {x}^{16}\right)$

Now multiplying and dividing by \left(1 - x\right)$both side we get,</p><p>\left(1 - x\right)$y \left(x\right)$= \left(1 - x\right) \left(1 + x\right)\left(1 + {x}^{2}\right)$\left(1 + {x}^{4}\right) \left(1 + {x}^{8}\right) \left(1 + {x}^{16}\right)$</p><p>$⇒ y\left(1 - x\right) = 1 - {x}^{32}$</p><p>$⇒ y - xy = 1 - {x}^{32} . . . . . . ( 1 )$</p><p>Now differentiating equation \left(1\right)$ on both sides with respect to $x$ we get,

${y}^{'} - x{y}^{'} - y = - 32{x}^{31} . . . . . . . ( 2 )$

Now at $x = - 1$

$⇒ {y}^{'} ( - 1 ) = 16$

Now differentiating equation \left(2\right)$with respect to$x$on both sides we get,</p><p>$⇒ {y}^{"} - x{y}^{"} - {y}^{'} - {y}^{'} = - \left(32\right)\left(31\right)${x}^{30}$

Now at $x = - 1$

⇒ {y}^{' '} ( - 1 ) = - 480

So, at $x = - 1$, ${y}^{'} - {y}^{"} = 16 - ( - 480 ) = 496$