What is the tangent of x^3+y^3=6xy,(3,3) ?

The tangent of x^3+y^3=6xy,(3,3) is y=-x+6

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tangent of x^3+y^3=6xy,(3,3)

tangent of

x^{3} + y^{3} = 6 x y, (3, 3)

y = - x + 6

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 3) : m = - 1

Find the line with slope m=

- 1

and passing through

(3, 3) : y = - x + 6

y = - x + 6

x - x^{3}

\int \frac{3 e ^{2 x}}{e ^{2 x} + 15 e ^{x} + 54} d x

f (x) = \frac{4}{x ^{2}}

\int \frac{t}{e ^{t^{2}}} d t

\int_{0}^{\infty} x^{3} e^{- x} d x

What is the tangent of x^3+y^3=6xy,(3,3) ?
The tangent of x^3+y^3=6xy,(3,3) is y=-x+6