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Question
Find `"dy"/"dx"` if, xy = log (xy)
Solution
xy = log (xy)
Differentiating both sides w.r.t. x, we get
`"x" * "dy"/"dx" + "y" * "d"/"dx" ("x") = 1/"xy" * "d"/"dx" ("xy")`
∴ `"x" * "dy"/"dx" + "y" = 1/"xy" ("x" "dy"/"dx" + "y") = 1/"y" "dy"/"dx" + 1/"x"`
∴ `("x" - 1/"y") "dy"/"dx" = 1/"x" - "y"`
∴ `- ((1 - "xy")/"y") "dy"/"dx" = ((1 - "xy")/"x")`
∴ `"dy"/"dx" = (- "y")/"x"`
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