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प्रश्न
If x = `"e"^(x/y)`, prove that `"dy"/"dx" = (x - y)/(xlogx)`
उत्तर
Given that: 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*1 - x* "dy"/"dx")/y^2`
⇒ y2 = `xy - x^2 * "dy"/"dx"`
⇒ `x^2 * "dy"/"dx"` = xy – y2
⇒ `"dy"/"dx" = (y(x - y))/x^2`
⇒ `"dy"/"d" = y/x * ((x - y)/x)`
⇒ `"dy"/"dx" = 1/logx * ((x - y)/x)` .....[∵ log x = `x/y` from equation (i)]
Hence, `"dy"/"dx" = (x - y)/(xlogx).
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