Question

If $y=\left(1 + x\right)\left(1 + x^{2}\right)\left(1 + x^{4}\right)\ldots \left(1 + x^{2^{n}}\right)$ , then $\frac{d y}{d x}$ at $x=0$ is:

Answer 1

Given, $y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)$
Taking the logarithm on both side of above equation (1), we get:
$ \begin{array}{l} \log _{e}(y)=\log _{e}\left((1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)\right) \\ \log _{e}(y)=\log _{e}(1+x)+\log _{e}\left(1+x^{2}\right)+\ldots+\log _{e}\left(1+x^{2^{n}}\right) \end{array} $
Differentiating equation (2) with respect to $x$, we get:
$\frac{d}{d x}\left(\log _{e}(y)\right)=\frac{d}{d x}\left[\log _{e}(1+x)+\log _{e}\left(1+x^{2}\right)+\ldots+\log _{e}\left(1+x^{2^{n}}\right)\right]$
$\frac{d}{d x}\left(\log _{e}(y)\right)=\frac{d}{d x}\left(\log _{e}(1+x)\right)+\frac{d}{d x}\left(\log _{e}\left(1+x^{2}\right)\right)+\ldots+\frac{d}{d x}\left(\log _{e}\left(1+x^{2^{n}}\right)\right)$
$\frac{1 d y}{y} \frac{d x}{d x}=\frac{1}{1+x}+\frac{1}{1+x^{2}}(2 x)+\ldots+\frac{1}{1+x^{2^{n}}}\left(2^{n} x^{2^{n}-1}\right)$
From equation (1), we have:
At $x=0, y=1$.
Put $x=0$ in equation (3), we get:
$ \left(\frac{d y}{d x}\right)_{x=0}=1 $
Alternative solution:
Multiply and divide by $(1-x)$
$ \begin{array}{l} y=\frac{(1-x)(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)}{(1-x)} \\ \text { Use: }(a+b)(a-b)=a^{2}-b^{2} \end{array} $
$y=\frac{1-x^{2^{n+1}}}{1-x}=1+x+x^{2}+\ldots+x^{2^{n+1}}-1$, using Sum of first $n$ terms of a G.P.:
$a\left(\frac{1-r^{n}}{1-r}\right)=a+a r+a r^{2}+\ldots+a r^{n-1}$
$ \frac{d y}{d x}=1+2 x+3 x^{2}+\ldots+\left(2^{n+1}-1\right) x^{2^{n+1}-2} $
At $x=0, \frac{d y}{d x}=1$