Question: If y = (1 + x) (1 + x2) (1 + x4) …. (1 + x2n), then \(\frac{dy}{dx}...

Question

If y = (1 + x) (1 + x²) (1 + x⁴) …. (1 + x²ⁿ), then $\frac{dy}{dx}$ at x = 0 is

Answer 1

y = $\frac{(1 - x)}{(1 - x)}$ (1 + x²)….(1 + x²ⁿ) = $\frac{1 - (x^{2n})^{2}}{1 - x}$

$\frac{dy}{dx} = \frac{(1 - x)(0 - 4n^{4n - 1}) - (1 - x^{2n})( - 1)}{(1 - x)^{2}}$ ; at x = 0 ⇒ 1

Question: If y = (1 + x) (1 + x<sup>2</sup>) (1 + x<sup>4</sup>) …. (1 + x<sup>2n</sup>), then \(\frac{dy}{dx}...