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

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

When $x = -1$:

The first factor $(x+1) = (-1+1) = 0$

So $y(-1) = 0$

When applying the product rule, any term containing $(x+1)$ will equal zero at $x = -1$.

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For a product of four functions, the derivative is:

$\frac{dy}{dx} = \frac{d}{dx}(x+1) \cdot (x^2+1)(x^4+1)(x^8+1)$

$+ (x+1) \cdot \frac{d}{dx}(x^2+1) \cdot (x^4+1)(x^8+1)$

$+ (x+1)(x^2+1) \cdot \frac{d}{dx}(x^4+1) \cdot (x^8+1)$

$+ (x+1)(x^2+1)(x^4+1) \cdot \frac{d}{dx}(x^8+1)$

The individual derivatives are:

$\frac{d}{dx}(x+1) = 1$

$\frac{d}{dx}(x^2+1) = 2x$

$\frac{d}{dx}(x^4+1) = 4x^3$

$\frac{d}{dx}(x^8+1) = 8x^7$

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At $x = -1$:

$\frac{dy}{dx}\bigg|_{x=-1} = (1) \cdot (2)(2)(2) + (0) \cdot (\ldots) + (0) \cdot (\ldots) + (0) \cdot (\ldots)$

For the first term:

$((-1)^2+1) = 2$

$((-1)^4+1) = 2$

$((-1)^8+1) = 2$

First term $= 1 \times 2 \times 2 \times 2 = 8$

All other terms contain the factor $(x+1)$, which equals $0$ at $x = -1$.

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$\frac{dy}{dx}\bigg|_{x=-1} = 8$