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Question
If `"x"^(5/3) . "y"^(2/3) = ("x + y")^(7/3)` , the show that `"dy"/"dx" = "y"/"x"`
Solution
`"x"^(5/3) . "y"^(2/3) = ("x + y")^(7/3)`
Taking logarithm of both the sides , we get
`5/3 "log x" + 2/3 "log y" = 7/3 "log (x + y)"`
Differentiating both sides w.r.t. x,
`5/3 . 1/"x" + 2/3 . 1/"y" "dy"/"dx" = 7/3 . 1/("x + y") [1 + "dy"/"dx"]`
`therefore 2/"3y" "dy"/"dx" - 7/(3 ("x + y")) "dy"/"dx" = 7/(3 ("x + y")) - 5/"3x"`
`therefore [2/"3y" - 7/(3 ("x + y"))] "dy"/"dx" = (7"x" - 5("x + y"))/("3x"("x + y"))`
`therefore [(2("x + y") - "7y")/("3y" ("x + y"))] "dy"/"dx" = (7"x" - 5("x + y"))/("3x"("x + y"))`
`therefore ("2x" - "5y")/"y" "dy"/"dx" = (2"x" - 5"y")/"x"`
`therefore "dy"/"dx" = "y"/"x"`
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