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Question
If `x = e^(x/y)`, then show that `"dy"/"dx" = (x - y)/(xlogx)`
Solution
`x = e^(x/y)`
∴ `x/y` = log x ...(1)
∴ y = `x/logx`
∴ `"dy"/"dx" = "d"/"dx"(x/logx)`
= `((logx)."d"/"dx"(x) - x."d"/"dx"(logx))/((logx)`
= `((logx) xx 1 - x xx (1)/x)/((logx)^2`
= `(logx - 1)/((logx)(logx)`
= `(x/y - 1)/((x/y)(logx)` ...[By (1)]
= `(x - y)/(xlogx)`.
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