What is the tangent of x^3+y^3-9xy=0,(2,4) ?

The tangent of x^3+y^3-9xy=0,(2,4) is y= 4/5 x+12/5

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tangent of x^3+y^3-9xy=0,(2,4)

tangent of

x^{3} + y^{3} - 9 x y = 0, (2, 4)

y = \frac{4}{5} x + \frac{12}{5}

Solution steps

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

\frac{4}{5}

and passing through

(2, 4) : y = \frac{4}{5} x + \frac{12}{5}

y = \frac{4}{5} x + \frac{12}{5}

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

x \to \frac{π}{4} lim (3 cos (x))

\frac{d}{d x} (x (8 - x)^{\frac{1}{3}})

\int - 3 e^{- 3 t} d t

What is the tangent of x^3+y^3-9xy=0,(2,4) ?
The tangent of x^3+y^3-9xy=0,(2,4) is y= 4/5 x+12/5