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Question
If `xsqrt(1 - y^2) + ysqrt(1 - x^2)` = 1, then show that `"dy"/"dx" = -sqrt((1 - y^2)/(1 - x^2)`.
Solution
`xsqrt(1 - y^2) + ysqrt(1 - x^2)` = 1
∴ `ysqrt(1 - x^2) + xsqrt(1 - y^2)` = 1
Differentiating both sides w.r.t. x, we get
`y."d"/"dx"(sqrt(1 - x^2)) + sqrt(1 - x^2)."dy"/"dx" + x."d"/"dx"(sqrt(1 - y^2)) + sqrt(1 - y^2)."d"/"dx"(x)` = 0
∴ `y xx (1)/(2sqrt(1 - x^2))."d"/"dx"(1 - x^2) + sqrt(1 - x^2)."dy"/"dx" + x xx (1)/(2sqrt(1 - y^2))."d"/"dx"(1 - y^2) + sqrt(1 - y^2) xx 1` = 0
∴ `y/(2sqrt(1 - x^2)) xx (0 - 2x) + sqrt(1 - x^2)."dy"/"dx" + x/(2sqrt(1 - y^2)) xx (0 - 2y"dy"/"dx") + sqrt(1 - y^2)` = 0
∴ `(-xy)/sqrt(1 - x^2) + sqrt(1 - x^2)."dy"/"dx" - "xy"/sqrt(1 - y^2)."dy"/"dx" + sqrt(1 - y^2)` = 0
∴ `(sqrt(1 - x^2) - "xy"/sqrt(1 - y^2))"dy"/"dx" = "xy"/sqrt(1 - x^2) - sqrt(1 - y^2)`
∴ `[(sqrt(1 - x^2).sqrt(1 - y^2) - xy)/sqrt(1 - y^2)]"dy"/"dx" = (xy - sqrt(1 - x^2).sqrt(1 - y^2))/sqrt(1 - x^2)`
∴ `(1)/sqrt(1 - y^2)."dy"/"dx" = (-1)/sqrt(1 - x^2)`
∴ `"dy"/"dx" = -sqrt((1 - y^2)/(1 - x^2)`
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