If $y=(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2n})$, then the value of $\frac{dy}{dx}$ at $x=0$ is:

Given $y=(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2n})$.Taking natural logarithm on both sides:$\ln y=\ln (1+x)+\ln (1+x^2)+\ln (1+x^4)+\cdots +\ln (1+x^{2n})$ Differentiating both sides with respect to $x$:$\frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots +\frac{2n{x}^{2n-1}}{1+x^{2n}}$$\frac{dy}{dx}=y\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\cdots +\frac{2n{x}^{2n-1}}{1+x^{2n}}\right]$ At $x=0$:$y(0)=(1+0)(1+0)\cdots (1+0)=1$ The term inside the bracket at $x=0$ is:$\left[\frac{1}{1+0}+\frac{0}{1+0}+\cdots +\frac{0}{1+0}\right]=1$ Therefore, $\frac{dy}{dx}{|}_{x=0}=1\times 1=1$.