If \(y = (x+1)(x^2 + 1)(x^4 + 1)(x^8 + 1)\) then \(\displaystyle\frac{dy}{dx}\) at \(x=-1\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
Let \(u = x + 1\) and \(v = (x^2 + 1)(x^4 + 1)(x^8 + 1)\), so that \(y = uv\). By the product rule:
\( \displaystyle \frac{dy}{dx} = v\,\frac{du}{dx} + u\,\frac{dv}{dx} = v + (x + 1)\frac{dv}{dx} \)
At \(x = -1\) the factor x + 1 is zero, so the second term vanishes:
\( \displaystyle \left.\frac{dy}{dx}\right|_{x=-1} = v(-1) \)
\(= (1 + 1)(1 + 1)(1 + 1) = 8\)