Find dy/dx..?

Question

y = lim(n~infinity) (1+x)(1+x^2)(1+x^4)(1+x^8)......(1+x^2n)

Find dy/dx?
Can anyone help...

As Ron W said the nth term is (1-X^2^n)
sorry for the typo. u guys have been great.

Answer 1

Let z = (1+x)(1+x^2)(1+x^4)(1+x^8).. (1 + x^2^n)
=> (1 - x) z = (1 - x) (1+x)(1+x^2)(1+x^4)(1+x^8)..... (1 + x^2^n)
=> (1 - x) z = (1 - x^2)(1+x^2)(1+x^4)(1+x^8)..... (1 + x^2^n)
=> (1 - x) z = (1- x^4)(1+x^4)(1+x^8)..... (1 + x^2^n)
------------------------------------------------------
-------------------------------------------------------
=> (1 - x) z = (1 - x^2^n)(1 + x^2^n)
=> z = {1 - x^2^(n+1)}/(1 - x)
=>y = lim(n ->infinity){1 - x^2^(n+1)}/(1 - x)
assuming that x is numerically less than 1
y = 1/(1 - x)
y = (1 - x)^-1
dy/dx = {-1(1 - x)^-2} (-1) (using chain rule)
= 1/(1 - x)^2

Answer 2

Just a comment...

The jth term of the product is (1 + x^(2^j)), not (1 + x^(2j)), for j = 0,1,2,...

The limit is 1/(1-x) for -1 < x < 1

Answer 3

I don't know how to simplify the answer, but when you differentiate you get

dy/dx = y/(1+x) + y*(2x)/(1+x^2) + y*(4x^3)/(1+x^4) + ...