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Question
If x7 . y9 = (x + y)16 then show that `"dy"/"dx" = "y"/"x"`
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Solution
Given x7.y9 =(x+y)16
Taking log on both sides
Log(x7.y9) = log(x+y)16
7 log x + 9 log y - 16 log (x+y)
Differentiating w.r.t.x
`7. 1/"x" + 9. 1/"y" "dy"/"dx" = 16 . 1/("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"`
`=>"dy"/"dx" [9/"y" - 16/"x + y"] = (16 "x" - 7 ("x + y"))/("x" ("x + y"))`
`=>"dy"/"dx" [(9"x" + 9"y" - 16"y")/("y"("x" + "y"))] = (16"x" - 7"x" - 7"y")/"x (x + y)"`
`=> "dy"/"dx" [(9"x" - 7"y")/("y" ("x + y"))] = (9"x" - 7"y")/"x (x +y)"`
`=> "dy"/"dx" = (9"x" - 7"y")/("x (x + y)") xx ("y" ("x + y"))/(9"x" - 7"y") = "y"/"x"`
`=> "dy"/"dx" = "y"/"x"`
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