If $y = (x+1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$ then $\frac{dy}{dx}$ at $x=-1$ is

8

The correct answer is Option (1) → 8

Given: $y = (x+1)(x^2+1)(x^4+1)(x^8+1)$

Use product rule to find $\frac{dy}{dx}$ and evaluate at $x = -1$

Let: $A = x+1$
$B = x^2 + 1$
$C = x^4 + 1$
$D = x^8 + 1$

Then $y = A \cdot B \cdot C \cdot D$

Differentiate using product rule:

$\frac{dy}{dx} = A'BCD + AB'CD + ABC'D + ABCD'$

Now compute each derivative:

$A' = 1$

$B' = 2x$

$C' = 4x^3$

$D' = 8x^7$

At $x = -1$:

$A = 0$, $B = 2$, $C = 2$, $D = 2$

So $A'BCD = 1 \cdot 2 \cdot 2 \cdot 2 = 8$

All other terms have factor $A = 0$ ⟹ they become 0

Answer: 8