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

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

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implicit (dy)/(dx),y^3=x+y

implicit derivative

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

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

Solution steps

y^{3} = x + y

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

\int x + \frac{1}{x} d x

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

(0, 2), (4, - 2)

\frac{x}{x ^{2} - 6}

\int \frac{1}{16 sin ^{2} ( x )} d x

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