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Question
Find `"dy"/"dx"` for the following function.
x3 + y3 + 3axy = 1
Solution
x3 + y3 + 3axy = 1
Differentiating both sides with respect to x,
`3x^2 + 3y^2 "dy"/"dx" + 3"a" (x "dy"/"dx" + y * 1) = 0`
`3[x^2 + y^2 "dy"/"dx" + "a"x "dy"/"dx" + "a"y] = 0`
`x^2 + y^2 "dy"/"dx" + "a"x "dy"/"dx" + "a"y = 0`
`"dy"/"dx" [y^2 + "a"x] = - x^2 - "a"y`
`"dy"/"dx" = (- x^2 - "a"y)/(y^2 + "a"x)`
`= - (x^2 + "a"y)/(y^2 + "a"x)`
`= - [(x^2 + "a"y)/(y^2 + "a"x)]`
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