Question

If `y=(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))` then `(dy)/(dx)` at `x=0` is

Answer 1

Correct Answer - 1
`(dy)/(dx)=1(1+x^(2))(1+x^(4))...(1+x^(2^(n)))+2x(1+x)(1+x^(4))...(1+x^(2^(n)))`
`+4x^(3)(1+x)(1+x^(2))(1+x^(8))...(1+x^(2^(n)))`
:
`+2^(n)x^(2^(n-1))(1+x)(1+x^(2))...(1+x^(2^(n-1)))`
`"or "(dy)/(dx):|_(x=0)=1`