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प्रश्न
If x = `e^(x/y)`, then prove that `dy/dx = (x - y)/(xlogx)`.
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उत्तर
Given: x = `e^(x/y)`
Taking log on both the sides,
log x = `log e^(x/y)`
⇒ log x = `x/y log e`
⇒ log x = `x/y` ...[∵ log e = 1] ...(i)
Differentiating both sides w.r.t. x:
`d/dx log x = d/dx (x/y)`
⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`
⇒ `y^2 = xy - x^2 xx dy/dx`
⇒ `x^2 xx dy/dx = xy - y^2`
⇒ `dy/dx = (y(x - y))/x^2`
⇒ `dy/d = y/x xx ((x - y)/x)`
⇒ `dy/dx = 1/logx xx ((x - y)/x) ...[∵ log x = x/y "from equation (i)"]`
`dy/dx = (x - y)/(xlogx)`
Hence proved.
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