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Question

If y=logx+logx+logx+... show that dydx=1x(2y1).

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Solution

We have,
y=logx+logx+logx.....
y=logx+y (squaring)
y2=logx+y
y2y=logx
On differentiating w.r.t x, we get
(2y1)dydx=1x
dydx=1x(2y1)
Hence proved.

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