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