What is the tangent of y^3+3y^2=x^4-3x^2,(-2,1) ?

The tangent of y^3+3y^2=x^4-3x^2,(-2,1) is y=-20/9 x-31/9

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tangent of y^3+3y^2=x^4-3x^2,(-2,1)

tangent of

y^{3} + 3 y^{2} = x^{4} - 3 x^{2}, (- 2, 1)

y = - \frac{20}{9} x - \frac{31}{9}

Solution steps

Find the slope of

y^{3} + 3 y^{2} = x^{4} - 3 x^{2} : \frac{d y}{d x} = \frac{4 x ^{3} - 6 x}{3 y ( 2 + y )}

Compute the slope of

y^{3} + 3 y^{2} = x^{4} - 3 x^{2}

(- 2, 1) : m = - \frac{20}{9}

Find the line with slope m=

- \frac{20}{9}

and passing through

(- 2, 1) : y = - \frac{20}{9} x - \frac{31}{9}

y = - \frac{20}{9} x - \frac{31}{9}

\int 3 x (x - 2) d x

\int \frac{1}{x 2 - x ^{2}} d x

\int \frac{cos ( x )}{1 - cos ( x )} d x

2 x e^{2 x}

n = 9 \sum \infty e^{7 - 4 n}

What is the tangent of y^3+3y^2=x^4-3x^2,(-2,1) ?
The tangent of y^3+3y^2=x^4-3x^2,(-2,1) is y=-20/9 x-31/9