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Question
If log (x + y) = log (xy) + a then show that, `"dy"/"dx" = (- "y"^2)/"x"^2`.
Solution
log (x + y) = log (xy) + a
∴ log (x + y) = log x + log y + a
Differentiating both sides w.r.t. x, we get
`1/("x + y")*"d"/"dx" ("x + y") = 1/"x" + 1/"y" * "dy"/"dx"`
∴ `1/("x + y") (1 + "dy"/"dx") = 1/"x" + 1/"y" * "dy"/"dx"`
∴ `"dy"/"dx" (1/"y" - 1/("x + y")) = 1/("x + y") - 1/"x"`
∴ `"dy"/"dx" ["x"/("y"("x + y"))] = (-"y")/("x"("x + y"))`
∴ `"dy"/"dx" = - "y"^2/"x"^2`
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