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Question
If `"x"^7*"y"^9 = ("x + y")^16`, then show that `"dy"/"dx" = "y"/"x"`
Solution
`"x"^7*"y"^9 = ("x + y")^16`
Taking logarithm of both sides, we get
log `"x"^7*"y"^9` = log `("x + y")^16`
∴ log `"x"^7 + log "y"^9 = 16 log ("x + y")`
∴ 7 log x + 9 log y = 16 log (x + y)
Differentiating both sides w.r.t. x, we get
`7(1/"x") + 9(1/"y") "dy"/"dx" = 16(1/("x + y")) "d"/"dx" ("x + y")`
∴ `7/"x" + 9/"y" "dy"/"dx" = 16/("x + y") (1 + "dy"/"dx")`
∴ `7/"x" + 9/"y" "dy"/"dx" = 16/("x + y") + 16/("x + y") "dy"/"dx"`
∴ `9/"y" "dy"/"dx" - 16/("x + y") "dy"/"dx" = 16/("x + y") - 7/"x"`
∴ `(9/"y" - 16/("x + y")) "dy"/"dx" = 16/("x + y") - 7/"x"`
∴ `[("9x" + "9y" - 16"y")/("y"("x + y"))] "dy"/"dx" = (16"x" - 7"x" - 7"y")/("x"("x + y"))`
∴ `[("9x" - 7"y")/("y"("x + y"))] "dy"/"dx" = ("9x" - 7"y")/("x"("x + y"))`
∴ `"dy"/"dx" = ("9x" - 7"y")/("x"("x + y")) xx ("y"("x + y"))/("9x" - 7"y")`
∴ `"dy"/"dx" = "y"/"x"`
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