What is the tangent of y^3-xy^2+x^3=5,(1,2) ?

The tangent of y^3-xy^2+x^3=5,(1,2) is y= 1/8 x+15/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y^3-xy^2+x^3=5,(1,2)

tangent of

y^{3} - x y^{2} + x^{3} = 5, (1, 2)

y = \frac{1}{8} x + \frac{15}{8}

Solution steps

Find the slope of

y^{3} - x y^{2} + x^{3} = 5 : \frac{d y}{d x} = \frac{y ^{2} - 3 x ^{2}}{y ( - 2 x + 3 y )}

Compute the slope of

y^{3} - x y^{2} + x^{3} = 5

(1, 2) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(1, 2) : y = \frac{1}{8} x + \frac{15}{8}

y = \frac{1}{8} x + \frac{15}{8}

\int x cosh (x) d x

f (x) = x^{3} cos (x)

\frac{d}{d x} ((sin (x))^{\frac{1}{3}})

24 e^{- 7 t} - 5 e^{- 23 t}

\frac{d y}{d t} - 2 y = 4 - t

What is the tangent of y^3-xy^2+x^3=5,(1,2) ?
The tangent of y^3-xy^2+x^3=5,(1,2) is y= 1/8 x+15/8