If $y= \frac1{1+x} + \frac2{1+x^2} + \frac4{1+x^4} + \dots + \frac{2^n}{1+x^{2^n}} $, then find $\frac{\mathrm dy}{\mathrm dx}$. 
If $y= \dfrac1{1+x} + \dfrac2{1+x^2} + \dfrac4{1+x^4} + \dots + \dfrac{2^n}{1+x^{2^n}} $, then find $\dfrac{\mathrm dy}{\mathrm dx}$.

I am stuck up in this question. I tried taking log on both sides and generate some simplified expression but was unable to do so. Any help would be highly appreciated. thanks

Hint:
$$\dfrac1{1+y}+\dfrac1{1-y}=\dfrac2{1-y^2}$$
Set $y=x,x^2,x^4,\cdots,x^{2^n}$ to find
$$y+\dfrac1{1+x}=\dfrac{2^n}{1-x^{2^{n+1}}}$$
