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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}, at x = 0

\frac{\partial}{\partial x} (- 4 cos (2 x))

\frac{d ^{2} y}{d x ^{2}} - 4 y = (x^{2} - 3) sin (2 x)

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

f (x) = 7 + 2 x + x^{3} (5 x^{2} + 1)^{5}

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