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प्रश्न
If xy - yx = ab, find `(dy)/(dx)`.
उत्तर
The given function is xy - yx = ab
Let xy = u and yx = v
Then , the function becomes u - v = ab
`(d"u")/(d"x") - (d"v")/(d"x") = 0` .....(1)
u = xy
⇒ log u = log(xy)
⇒ log u = y log x
Differentiating both sides with respect to x, we obtain
`(1)/("u") (d"u")/(d"x") = log "x" (d"y")/(d"x") + "y". (d)/(d"x") (log "x")`
⇒ `(d"u")/(d"x") = [ log "x" (d"y")/(d"x") + "y" .(1)/("x")]`
⇒ `(d"u")/(d"x") = "x"^"y" (log "x" (d"y")/(d"x") + ("y")/("x"))` ...(2)
`"v" = "y"^"x"`
⇒ `log "v" = log ("y"^"x")`
⇒ `log "v" = "x" log "y"`
Differentiating both sides with respect to x, we obtain
`(1)/("v").(d"v")/(d"x") = log "y".(d)/(d"x") ("x") + "x". (d)/(d"x") (log "y")`
⇒ `(d"v")/(d"x") = "v" (log "y". 1 + "x".(1)/("y"). (d"y")/(d"x"))`
⇒ `(d"v")/(d"x") = "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x"))` ...(3)
From (1), (2), and (3), we obtain
`"x"^"y" (log "x"(d"y")/(d"x") + ("y")/("x")) - "y"^"x" (log "y" + ("x")/("y") (d"y")/(d"x")) = 0`
⇒ `"x"^"y" log "x" (d"y")/(d"x") - "xy"^("x"-1) (d"y")/(d"x") + "x"^("y"-1) "y"-"y"^"x" log"y" = 0`
⇒ `("x"^"y" log "x" - "xy"^("x"-1)) (d"y")/(d"x") = "y"^"x" log "y" - "x"^("y"-1) "y"`
⇒ `(d"y")/(d"x") = ("y"^"x" log "y" - "x"^("y" -1) "y")/(("x"^"y" log "x" - "xy"^("x"-1))`
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