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Question
If `x^y = e^(x - y)` , show that `(dy)/(dx) = logx/(1 + logx)^2`
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Solution
`x^y = e^(x - y)`
Taking log on both sides ,
`log(x)^y = loge^(x - y)`
∴ y . logx = (x - y).loge
∴ y logx = x - y ...........(∵ loge = 1)
∴ y logx + y = x
∴ y(logx + 1) = x
∴ y = `x/(1 + logx)`
Differentiating w.r.t. x,
`(dy)/(dx) = ((1 + logx). 1 - x(0 + 1/x))/(1 + logx)^2`
∴ `(dy)/(dx) = (1 + logx - 1)/(1 + logx)^2`
∴ `(dy)/(dx) = logx/(1 + logx)^2`
Hence proved .
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