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Q.

If x=ey+eyey.....∞,x=0 then dydx=1−kxx,k= ?

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Detailed Solution

x=ey+xlog⁡x=log⁡ey+xlog⁡x=(y+x)log⁡elog⁡x=y+x1x=dydx+1dydx=1x−1=1−xx∴k=1
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If x=ey+eyey.....∞,x=0 then dydx=1−kxx,k= ?