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Question
If `"x"^3 + "y"^2 + "xy" = 7` Find `"dy"/"dx"`
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Solution
`"x"^3 + "y"^2 + "xy" = 7`
Differentiating both sides w.r.t. x, we get
`"3x"^2 + 2"y" "dy"/"dx" + "x""dy"/"dx" + "y""d"/"dx" "x" = 0`
∴ `"3x"^2 + 2"y" "dy"/"dx" + "x""dy"/"dx" + "y"(1) = 0`
∴ `2"y" "dy"/"dx" + "x""dy"/"dx" = - "y" - 3"x"^2`
∴ `(2"y" + "x") "dy"/"dx" = - "y" - 3"x"^2`
∴ `"dy"/"dx" = - ("y" + 3"x"^2)/("2y" + "x")`
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