What is the tangent of 2x^2+3xy+1y^3=-10,(-3,-4) ?

The tangent of 2x^2+3xy+1y^3=-10,(-3,-4) is y= 8/13 x-28/13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 2x^2+3xy+1y^3=-10,(-3,-4)

tangent of

2 x^{2} + 3 x y + 1 y^{3} = - 10, (- 3, - 4)

y = \frac{8}{13} x - \frac{28}{13}

Solution steps

Find the slope of

2 x^{2} + 3 x y + 1 y^{3} = - 10 : \frac{d y}{d x} = \frac{- 4 x - 3 y}{3 ( y ^{2} + x )}

Compute the slope of

2 x^{2} + 3 x y + 1 y^{3} = - 10

(- 3, - 4) : m = \frac{8}{13}

Find the line with slope m=

\frac{8}{13}

and passing through

(- 3, - 4) : y = \frac{8}{13} x - \frac{28}{13}

y = \frac{8}{13} x - \frac{28}{13}

x^{2} y^{^{''}} + x y^{^{'}} + y = sec (ln (x))

\frac{d}{d x} (\frac{x + 2}{x - 3})

\frac{d}{d x} (ln^{3} (3 x^{2} - 5))

\int (x^{2} + 2)^{3} d x

x \to 0 lim ((x - 4)^{3})

What is the tangent of 2x^2+3xy+1y^3=-10,(-3,-4) ?
The tangent of 2x^2+3xy+1y^3=-10,(-3,-4) is y= 8/13 x-28/13