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Question
If `"x"^"a"*"y"^"b" = ("x + y")^("a + b")`, then show that `"dy"/"dx" = "y"/"x"`
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Solution
`"x"^"a"*"y"^"b" = ("x + y")^("a + b")`
Taking logarithm of both sides, we get
log (`"x"^"a"*"y"^"b"`) = log `("x + y")^("a + b")`
∴ log `"x"^"a" + log "y"^"b" = ("a + b") log ("x + y")`
∴ a log x + b log y = (a + b) log (x + y)
Differentiating both sides w.r.t. x, we get
`"a"(1/"x") + "b"(1/"y") "dy"/"dx" = ("a + b")(1/("x + y")) "d"/"dx" ("x + y")`
∴ `"a"/"x" + "b"/"y" "dy"/"dx" = ("a + b")/("x + y") (1 + "dy"/"dx")`
∴ `"a"/"x" + "b"/"y" "dy"/"dx" = ("a + b")/("x + y") + ("a + b")/("x + y") "dy"/"dx"`
∴ `"b"/"y" "dy"/"dx" - ("a + b")/("x + y") "dy"/"dx" = ("a + b")/("x + y") - "a"/"x"`
∴ `("b"/"y" - ("a + b")/("x + y")) "dy"/"dx" = ("a + b")/("x + y") - "a"/"x"`
∴ `[("bx" + "by" - "a""y" - "by")/("y"("x + y"))] "dy"/"dx" = ("ax" + "bx" - "ax" - "ay")/("x"("x + y"))`
∴ `[("bx" - "ay")/("y"("x + y"))] "dy"/"dx" = ("bx" - "ay")/("x"("x + y"))`
∴ `"dy"/"dx" = ("bx" - "ay")/("x"("x + y")) xx ("y"("x + y"))/("bx" - "ay")`
∴ `"dy"/"dx" = "y"/"x"`
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