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