What is the tangent of 3x^3 ?

The tangent of 3x^3 is y=9a_{0}^2x-6a_{0}^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 3x^3

tangent of

3 x^{3}

y = 9 a_{0}^{2} x - 6 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 3 x^{3}

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

Find the line with slope m=

9 a_{0}^{2}

and passing through

(a_{0}, 3 a_{0}^{3}) : y = 9 a_{0}^{2} x - 6 a_{0}^{3}

y = 9 a_{0}^{2} x - 6 a_{0}^{3}

y^{^{'}} = x ln (x)

\frac{d}{d x} (\frac{8 x}{5})

\frac{d}{d x} ((3 x^{6} + 2 x^{3})^{4})

\int (2 x + 5) (x^{2} + 5 x) 7 d x

\int_{1}^{n} x ln (x) d x