Question

If y = (1 + x) (1 + x²) (1 + x⁴) …….. $(1 + x^{2^{n}})$ then $\frac{dy}{dx}$

at x = 0 is

• A. 1
• B. – 1
• C. 0
• D. None

Answer 1

Answer: (A)

1

Answer 2

$y = (1 + x)(1 + x^{2})(1 + x^{4})..........(1 + x^{2^{n}})$

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

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

= $\frac{1 - x^{2^{n + 1}}}{1 - x}$

$\frac{dy}{dx} = \frac{(1 - x)( - 2^{n + 1} \cdot x^{2^{n + 1} - 1}) + (1 - x^{2^{n + 1}})}{(1 - x)^{2}}$

$\left( \frac{dy}{dx} \right)_{(x = 0)} = \frac{1}{1} = 1$