What is the tangent of x^3+9x ?

The tangent of x^3+9x is y=(3a_{0}^2+9)x-2a_{0}^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^3+9x

tangent of

x^{3} + 9 x

y = (3 a_{0}^{2} + 9) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 9 a_{0})

Find the slope of

y = x^{3} + 9 x : \frac{d y}{d x} = 3 x^{2} + 9

Compute the slope of

y = x^{3} + 9 x

x = a_{0} : m = 3 a_{0}^{2} + 9

Find the line with slope m=

3 a_{0}^{2} + 9

and passing through

(a_{0}, a_{0}^{3} + 9 a_{0}) : y = (3 a_{0}^{2} + 9) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} + 9) x - 2 a_{0}^{3}

\frac{d}{d x} (\frac{2 x}{2 x + 1})

f (x) = ln (1 - 3 x)

\int_{0}^{N} e^{(- s t)} sin (5 t) d t

\frac{d}{d x} ((2 x^{2})^{\frac{1}{2}})

What is the tangent of x^3+9x ?
The tangent of x^3+9x is y=(3a_{0}^2+9)x-2a_{0}^3