What is the tangent of x^3y^2=xy^3+6,(2,1) ?

The tangent of x^3y^2=xy^3+6,(2,1) is y=-11/10 x+16/5

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tangent of x^3y^2=xy^3+6,(2,1)

tangent of

x^{3} y^{2} = x y^{3} + 6, (2, 1)

y = - \frac{11}{10} x + \frac{16}{5}

Solution steps

Find the slope of

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

Compute the slope of

x^{3} y^{2} = x y^{3} + 6

(2, 1) : m = - \frac{11}{10}

Find the line with slope m=

- \frac{11}{10}

and passing through

(2, 1) : y = - \frac{11}{10} x + \frac{16}{5}

y = - \frac{11}{10} x + \frac{16}{5}

\frac{d}{d x} (lo g_{5} (6 x^{3}))

y = \frac{1}{x - 4}, at x = 9

\int 7 e^{2} d x

\int cos^{3} (9 x) d x

\int_{0}^{1} \frac{1}{( x ^{2} + 1 )} d x

What is the tangent of x^3y^2=xy^3+6,(2,1) ?
The tangent of x^3y^2=xy^3+6,(2,1) is y=-11/10 x+16/5