Question

Find `"dy"/"dx"` if, x³ + y³ + 4x³y = 0 

Answer 1

x³ + y³ + 4x³y = 0 

Differentiating both sides w.r.t. x, we get,

`"d"/"dx" "x"^3 + "d"/"dx" "y"^3 + 4"d"/"dx" ("x"^3."y") = 0`

`3"x"^2 "d"/"dx" "x" + 3"y"^2 "dy"/"dx" + 4 ["x"^3 "dy"/"dx" + "y" "d"/"dx" ("x"^3)]` = 0

∴ `3"x"^2. (1) + 3"y"^2 "dy"/"dx" + 4 ["x"^3 "dy"/"dx" + "y"(3"x"^2)] = 0`

∴ `3"x"^2 + 3"y"^2 "dy"/"dx" + 4"x"^3 "dy"/"dx" + 12"x"^2"y" = 0`

∴ `(3"y"^2 + 4"x"^3) "dy"/"dx" = − (12"x"^2"y" + 3"x"^2)`

∴ `"dy"/"dx" = (- (12"x"^2"y" + 3"x"^2))/((3"y"^2 + 4"x"^3)) = - (3"x"^2(1 + 4"y"))/(3"y"^2 + 4"x"^3)`