What is the tangent of 2x^3+2xy+2y^2=38,(-1,-4) ?

The tangent of 2x^3+2xy+2y^2=38,(-1,-4) is y=-1/9 x-37/9

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tangent of 2x^3+2xy+2y^2=38,(-1,-4)

tangent of

2 x^{3} + 2 x y + 2 y^{2} = 38, (- 1, - 4)

y = - \frac{1}{9} x - \frac{37}{9}

Solution steps

Find the slope of

2 x^{3} + 2 x y + 2 y^{2} = 38 : \frac{d y}{d x} = - \frac{3 x ^{2} + y}{x + 2 y}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

(- 1, - 4) : y = - \frac{1}{9} x - \frac{37}{9}

y = - \frac{1}{9} x - \frac{37}{9}

x \to 0 lim (\frac{sin ( x )}{x} - 1)

\int_{2}^{5} \frac{1}{x - 2} d x

f (x) = (2 x^{3} + 5 x) (x - 3) (x + 2)

x \to π lim (3 cot (x))

\int \frac{15 e ^{5 r}}{e ^{5 r} + 2} d r

What is the tangent of 2x^3+2xy+2y^2=38,(-1,-4) ?
The tangent of 2x^3+2xy+2y^2=38,(-1,-4) is y=-1/9 x-37/9