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