Question

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

विकल्प

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Answer 1

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