What is implicit (dy)/(dx),x^3+y^3-9xy=0 ?

The answer to implicit (dy)/(dx),x^3+y^3-9xy=0 is (dy)/(dx)=(-x^2+3y)/(y^2-3x)

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implicit (dy)/(dx),x^3+y^3-9xy=0

implicit derivative

\frac{d y}{d x}, x^{3} + y^{3} - 9 x y = 0

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

Solution steps

x^{3} + y^{3} - 9 x y = 0

Treat

y

y (x)

Differentiate both sides:

3 x^{2} + 3 y^{2} \frac{d y}{d x} - 9 (y + x \frac{d y}{d x}) = 0

Isolate

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

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

f (t) = 1 + e^{4 t}

y = 2 x, (18, 6)

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

\int - 6 y d y

\int \frac{x}{1 + cos ( x )} d x

What is implicit (dy)/(dx),x^3+y^3-9xy=0 ?
The answer to implicit (dy)/(dx),x^3+y^3-9xy=0 is (dy)/(dx)=(-x^2+3y)/(y^2-3x)