What is implicit x^3+y^3=1 ?

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

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implicit x^3+y^3=1

implicit derivative

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

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

Solution steps

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

Treat

y

y (x)

Differentiate both sides:

3 x^{2} + 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}}

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

\int \frac{9 x ^{2}}{x ^{2} - 4} d x

x y^{^{'}} = y^{2} - y

y = \frac{x}{x - 5}

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

\int_{0}^{7} x^{2} - 49 d x

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