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If xxyyx1+x+y1+y = 0, x ≠ y, then xdydx(1+x)2dydx = ____________.

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Question

If `"x"/sqrt(1 + "x") + "y"/sqrt(1 + "y")` = 0, x ≠ y, then `(1 + "x")^2 "dy"/"dx"` = ____________.

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Solution

If `"x"/sqrt(1 + "x") + "y"/sqrt(1 + "y")` = 0, x ≠ y, then `(1 + "x")^2 "dy"/"dx"` = −1

Explanation:

We have, `"x"/sqrt(1 + "x") + "y"/sqrt(1 + "y")` = 0

`"x" sqrt(1 + "y") + "y" sqrt(1 + "x")` = 0

`"x" sqrt(1 + "y") = -"y" sqrt(1 + "x")`

x2(1 + y) = y2(1 + x)

x2 + x2y − y2 − xy2 = 0

(x2 − y2) + x2y − xy2 = 0

(x + y) (x − y) + xy(x − y) = 0

(x − y) (x + y + xy) = 0

y(1 + x) = −x ...............[∵ x − y ≠ 0]

y = `(-"x")/(1 + "x")`

`"dy"/"dx" = - [((1 + "x") (1) - ("x"))/(1 + "x")]^2`

`(1 + "x")^2 "dy"/"dx"` = −1

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Derivative of Implicit Functions
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