हिंदी

If xy = ex–y, then show that dydxdydx=logx(1+logx)2.

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प्रश्न

If xy = ex–y, then show that `"dy"/"dx" = logx/(1 + logx)^2`.

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उत्तर

xy = ex–y  

∴ log xy = log ex-y    

∴ y log x = (x – y) log e

∴ y log x = x – y     ...[∵ log e = 1]

∴ y + y log x = x        ∴ y(1 + log x) = x

∴ y = `x/(1 + log x)`

∴ `"dy"/"dx" = "d"/"dx"(x/(1 + log x))`

= `((1 + log x)."d"/"dx"(x) - x"d"/"dx"(1 + log x))/(1 + log x)^2`

= `((1 + log x).1 - x(0 + 1/x))/(1 + logx)^2`

= `(1 + logx - 1)/(1 + log x)^2`

= `log x/(1 + log x)^2`.

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अध्याय 1: Differentiation - Exercise 1.3 [पृष्ठ ४०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.3 | Q 5.06 | पृष्ठ ४०

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