What is the tangent of 2x^3+6x^2-90x-6 ?

The tangent of 2x^3+6x^2-90x-6 is y=(6a_{0}^2+12a_{0}-90)x-4a_{0}^3-6a_{0}^2-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 2x^3+6x^2-90x-6

tangent of

2 x^{3} + 6 x^{2} - 90 x - 6

y = (6 a_{0}^{2} + 12 a_{0} - 90) x - 4 a_{0}^{3} - 6 a_{0}^{2} - 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 6 a_{0}^{2} - 90 a_{0} - 6)

Find the slope of

y = 2 x^{3} + 6 x^{2} - 90 x - 6 : \frac{d y}{d x} = 6 x^{2} + 12 x - 90

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6 a_{0}^{2} + 12 a_{0} - 90

Find the line with slope m=

6 a_{0}^{2} + 12 a_{0} - 90

and passing through

(a_{0}, 2 a_{0}^{3} + 6 a_{0}^{2} - 90 a_{0} - 6) : y = (6 a_{0}^{2} + 12 a_{0} - 90) x - 4 a_{0}^{3} - 6 a_{0}^{2} - 6

y = (6 a_{0}^{2} + 12 a_{0} - 90) x - 4 a_{0}^{3} - 6 a_{0}^{2} - 6

\int 4 sin (x) d x

x \cdot 1 - x^{2}

h \to 0 lim \frac{( 2 ( 3 + h ) ^{4} - 162 )}{h}

\frac{d}{d x} (3 ln (7 x))

\int - e^{3 t} d t

What is the tangent of 2x^3+6x^2-90x-6 ?
The tangent of 2x^3+6x^2-90x-6 is y=(6a_{0}^2+12a_{0}-90)x-4a_{0}^3-6a_{0}^2-6