Question

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

Answer 1

Answer: 1

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