Question

If $y=(1+x)(1+x^2)(1+x^4)\dots (1+x^{2^n}),$ then $y^{\prime }\left(0\right)$ is equal to

Answer 1

$\ln y=\ln (1+x)+\ln (1+x^2)+\ln (1+x^4)+\cdots +\ln (1+x^{2^n})$
differentiating w.r.t. $x$ both sides,
$\Rightarrow \frac{y^{\prime }}{y}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots +\frac{2^n{x}^{2^n-1}}{(1+x^{2^n})}\Rightarrow y^{\prime }\left(0\right)=1,(\because y(0)=1)$