For |x|<1,let y=1+x+x2+… to ∞, then dydxis equal to 

For |x|<1,let y=1+x+x2+ to ∞, then dydxis equal to 

  1. A

    xy

  2. B

    x2y2

  3. C

    xy2

  4. D

    xy2+y

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    Solution:

    y=1+x+x2+ y=11x=(1x)1

    On differentiating w.r.t. x, we get

    dydx=1(1x)2(1)=1(1x)2 dydxy=1(1x)21(1x)=11+x(1x)2=x(1x)2 dydxy=xy2dydx=xy2+y

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