What is the tangent of-3x^3-3x^2-x+3 ?

The tangent of-3x^3-3x^2-x+3 is y=(-9a_{0}^2-6a_{0}-1)x+6a_{0}^3+3a_{0}^2+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of-3x^3-3x^2-x+3

tangent of

- 3 x^{3} - 3 x^{2} - x + 3

y = (- 9 a_{0}^{2} - 6 a_{0} - 1) x + 6 a_{0}^{3} + 3 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 a_{0}^{3} - 3 a_{0}^{2} - a_{0} + 3)

Find the slope of

y = - 3 x^{3} - 3 x^{2} - x + 3 : \frac{d y}{d x} = - 9 x^{2} - 6 x - 1

Compute the slope of

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

x = a_{0} : m = - 9 a_{0}^{2} - 6 a_{0} - 1

Find the line with slope m=

- 9 a_{0}^{2} - 6 a_{0} - 1

and passing through

(a_{0}, - 3 a_{0}^{3} - 3 a_{0}^{2} - a_{0} + 3) : y = (- 9 a_{0}^{2} - 6 a_{0} - 1) x + 6 a_{0}^{3} + 3 a_{0}^{2} + 3

y = (- 9 a_{0}^{2} - 6 a_{0} - 1) x + 6 a_{0}^{3} + 3 a_{0}^{2} + 3

\frac{d}{d x} (\frac{1}{x ln ( x ^{9} )})

y^{^{'}} = \frac{y}{x} + \frac{x}{y}

y^{^{''}} + y^{^{'}} = sin (20 x)

\frac{d y}{d x} = \frac{1}{10} y cos^{2} (y)

\frac{d}{d x} (\frac{x ^{2} + 32 x}{8})

What is the tangent of-3x^3-3x^2-x+3 ?
The tangent of-3x^3-3x^2-x+3 is y=(-9a_{0}^2-6a_{0}-1)x+6a_{0}^3+3a_{0}^2+3