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Question
If `"x"^5 * "y"^7 = ("x + y")^12` then show that, `"dy"/"dx" = "y"/"x"`
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Solution
`"x"^5 * "y"^7 = ("x + y")^12`
Taking logarithm of both sides, we get
`log ("x"^5 * "y"^7) = log ("x + y")^12`
∴ log x5 + log y7 = 12 log (x + y)
∴ 5 log x + 7 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get
`5. 1/"x" + 7. 1/"y" * "dy"/"dx" = 12 * 1/("x + y") * "d"/"dx" ("x + y")`
∴ `5/"x" + 7/"y" * "dy"/"dx" = 12/("x + y") [1 + "dy"/"dx"]`
∴ `5/"x" + 7/"y" * "dy"/"dx" = 12/("x + y") + 12/("x + y") * "dy"/"dx"`
∴ `[7/"y" - 12/("x + y")] "dy"/"dx" = 12/("x + y") - 5/"x"`
∴ `[(7"x" + 7"y" - 12"y")/("y" ("x + y"))] "dy"/"dx" = (12"x" - 5"x" - 5"y")/("x"("x + y"))`
∴ `[("7x" - 5"y")/("y"("x + y"))] "dy"/"dx" = [("7x" - 5"y")/("x"("x + y"))]`
∴ `"dy"/"dx" = [("7x" - 5"y")/("x"("x + y"))] xx ("y"("x + y"))/("7x" - 5"y")`
∴ `"dy"/"dx" = "y"/"x"`
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