Question

If $y=(1+x)(1+x^2)(1+x^4)\dots (1+x^{2n})$ then the value of ${\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)}_{x=0}$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

$y=(1+x)(1+x^2)(1+x^4)\dots (1+x^{2n})$
Taking log on both sides
$logy=log(1+x)+log(1+x^2)+log(1+.x^4)+\cdots +log(1+x^{2n})$
On differentiating w.r.t x
$\frac{dy}{dx}=y\left[\begin{array}{c}\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\dots \end{array}\right]$
At $x=0,\frac{dy}{dx}=1$
Since, at $x=0,y=1$