What is implicit (dy)/(dx),x^3-3xy^2+y^3=1 ?

The answer to implicit (dy)/(dx),x^3-3xy^2+y^3=1 is (dy)/(dx)=(-x^2+y^2)/(y(y-2x))

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implicit (dy)/(dx),x^3-3xy^2+y^3=1

implicit derivative

\frac{d y}{d x}, x^{3} - 3 x y^{2} + y^{3} = 1

\frac{d y}{d x} = \frac{- x ^{2} + y ^{2}}{y ( y - 2 x )}

Solution steps

x^{3} - 3 x y^{2} + y^{3} = 1

Treat

y

y (x)

Differentiate both sides:

3 x^{2} - 3 (y^{2} + 2 x y \frac{d y}{d x}) + 3 y^{2} \frac{d y}{d x} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- x ^{2} + y ^{2}}{y ( y - 2 x )}

\frac{d y}{d x} = \frac{- x ^{2} + y ^{2}}{y ( y - 2 x )}

y^{^{'}} = 3 x + y, y (0) = 5

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 13} d x

y = 2 x, (1, 2)

y = \frac{x ^{4}}{16}

\int x (3 x - 5)^{10} d x

What is implicit (dy)/(dx),x^3-3xy^2+y^3=1 ?
The answer to implicit (dy)/(dx),x^3-3xy^2+y^3=1 is (dy)/(dx)=(-x^2+y^2)/(y(y-2x))