What is the tangent of x^3+y^3=8xy,(4,4) ?

The tangent of x^3+y^3=8xy,(4,4) is y=-x+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^3+y^3=8xy,(4,4)

tangent of

x^{3} + y^{3} = 8 x y, (4, 4)

y = - x + 8

Solution steps

Find the slope of

x^{3} + y^{3} = 8 x y : \frac{d y}{d x} = \frac{8 y - 3 x ^{2}}{3 y ^{2} - 8 x}

Compute the slope of

x^{3} + y^{3} = 8 x y

(4, 4) : m = - 1

Find the line with slope m=

- 1

and passing through

(4, 4) : y = - x + 8

y = - x + 8

x \to 3 lim (1 0^{x^{2} - 3 \cdot x})

\frac{\partial}{\partial y} (x e^{9 x y})

x + 2, x^{2} - 4

\frac{\partial}{\partial x} (3 y^{2})

\int e^{r} (2 + 3 e^{r})^{\frac{3}{2}} d r

What is the tangent of x^3+y^3=8xy,(4,4) ?
The tangent of x^3+y^3=8xy,(4,4) is y=-x+8