What is the slope of x^3+y^3-6xy=0,(4/3 , 8/3) ?

The slope of x^3+y^3-6xy=0,(4/3 , 8/3) is 4/5

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slope of x^3+y^3-6xy=0,(4/3 , 8/3)

slope

x^{3} + y^{3} - 6 x y = 0, (\frac{4}{3}, \frac{8}{3})

\frac{4}{5}

Solution steps

x^{3} + y^{3} - 6 x y = 0

Use implicit differentiation to find the slope of

x^{3} + y^{3} - 6 x y = 0

Implicit Derivative

\frac{d y}{d x}

x^{3} + y^{3} - 6 x y = 0 : \frac{- x ^{2} + 2 y}{y ^{2} - 2 x}

\frac{- x ^{2} + 2 y}{y ^{2} - 2 x}

Find the slope at

(\frac{4}{3}, \frac{8}{3}) : \frac{4}{5}

\frac{4}{5}

\int_{2}^{5} 4 x^{3} d x

f (x) = \frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}}

x \to c lim (327)

\int_{- \infty}^{- 1} e^{- 16 t} d t

\int x^{- \frac{1}{2}} ln (x) d x

What is the slope of x^3+y^3-6xy=0,(4/3 , 8/3) ?
The slope of x^3+y^3-6xy=0,(4/3 , 8/3) is 4/5