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

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

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

tangent of

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

y = - x + 4

Solution steps

Find the slope of

x^{3} + y^{3} - 4 x y = 0 : \frac{d y}{d x} = \frac{- 3 x ^{2} + 4 y}{3 y ^{2} - 4 x}

Compute the slope of

x^{3} + y^{3} - 4 x y = 0

(2, 2) : m = - 1

Find the line with slope m=

- 1

and passing through

(2, 2) : y = - x + 4

y = - x + 4

G (t) = \frac{1 - 4 t}{3 + t}

f (x) = \frac{sin ( x )}{x ^{2}}

\int x^{2} + 5 x + 5 d x

y y^{^{'}} - 2 e^{x} = 0

n \to \infty lim (sin (n))

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