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рдкреНрд░рд╢реНрди
If x = `e^(x/y)`, then show that `dy/dx = (x - y)/(xlogx)`
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рдЙрддреНрддрд░
x = `e^(x/y)`
∴ `x/y` = log x ...(1)
∴ y = `x/logx`
∴ `dy/dx = d/dx(x/log x)`
= `((log x) * d/dx(x) - x * d/dx(log x))/((log x)`
= `((log x) xx 1 - x xx (1)/x)/((log x)^2`
= `(log x - 1)/((log x)(log x)`
= `(x/y - 1)/((x/y)(log x)` ...[By (1)]
= `(x - y)/(x log x)`
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