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प्रश्न
xy = ex-y, then show that `"dy"/"dx" = ("log x")/("1 + log x")^2`
उत्तर
xy = ex-y
Taking logarithm of both the sides
y log x = (x - y) log e
= x - y ...(1)
Diff. both the sides w.r.t. x ,
`"y".1/"x" + "log x" . "dy"/"dx" = 1 - "dy"/"dx"`
`therefore "log x" . "dy"/"dx" + "dy"/"dx" = 1 - "y"/"x"`
`therefore (1 + "log x") "dy"/"dx" = ("x - y")/"x"`
`therefore (1 + "log x") "dy"/dx"= (ylogx)/x` ...[by (1)]
`"dy"/"dx" = ("y log x")/("x" . (1 + "log x"))`
`= ("log x")/(1 + "log x")^2`
`[because "from (1)" "y"/"x" = 1/(1 + "log x")]`
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