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Question
If log (x + y) = log(xy) + p, where p is a constant, then prove that `"dy"/"dx" = (-y^2)/(x^2)`.
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Solution
log (x + y) = log(xy) + p
∴ log( x + y) = logx + logy + p
Differentiating both sides w.r.t. x, we get
`(1)/(x + y)."d"/"dx"(x + y) = (1)/x + (1)/y."dy"/"dx" + 0`
∴ `(1)/(x + y)(1 + "dy"/"dx") = (1)/x + (1)/y."dy"/"dx"`
∴ `(1)/(x + y) + (1)/(x + y)."dy"/"dx" = (1)/x + (1)/y."dy"/"dx"`
∴ `(1/(x + y) - 1/y)"dy"/"dx" = (1)/x - (1)/(x + y)`
∴ `[(y - x - y)/(y(x + y))]"dy"/"dx" = (x + y - x)/(x(x + y)`
∴ `[(-x)/(y(x + y))]"dy"/"dx" = y/(x(x + y)`
∴ `(-x/y)"dy"/"dx" = y/x`
∴ `"dy"/"dx" = -y^2/x^2`.
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