What is the tangent of x^3,\at x=2 ?

The tangent of x^3,\at x=2 is y=12x-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^3,\at x=2

tangent of

x^{3}, at x = 2

y = 12 x - 16

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

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

Compute the slope of

y = x^{3}

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

\int (x - 1)^{2} \cdot x^{3} \cdot (2 - x) d x

\frac{d y}{d x} = \frac{5 x}{9 y}

\frac{d}{d x} (7 e^{e})

\frac{\partial}{\partial x} ((x^{3} + y^{2})^{\frac{1}{2}})

y^{^{''}} + 2 y^{^{'}} + 2 y = 1