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

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

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

tangent of

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

y = \frac{5}{4} x - 3

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

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

Find the line with slope m=

\frac{5}{4}

and passing through

(4, 2) : y = \frac{5}{4} x - 3

y = \frac{5}{4} x - 3

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

\int \frac{( 4 x ^{m} - 3 x ^{n} ) ^{3}}{x} d x

x \to \infty lim ((- 1)^{- x})

\int (x^{4} - x^{2} + \frac{1}{4 x ^{3}} - \frac{1}{x ^{2}}) d x

\frac{3}{( s ^{2} + 4 )}

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