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

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

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

tangent of

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

y = \frac{4}{5} x + \frac{8}{5}

Solution steps

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

\frac{4}{5}

and passing through

(\frac{4}{3}, \frac{8}{3}) : y = \frac{4}{5} x + \frac{8}{5}

y = \frac{4}{5} x + \frac{8}{5}

\frac{\partial}{\partial x} (e^{x^{2} y} + x y^{3})

\frac{d}{d x} (9 ln (4 x))

x - 4

\frac{d}{d x} (e^{x^{y}})

\frac{d}{d x} (e^{x^{2}} sin (x^{3} - 2))

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