What is the tangent of-9x^4+x^3 ?

The tangent of-9x^4+x^3 is y=(-36a_{0}^3+3a_{0}^2)x+27a_{0}^4-2a_{0}^3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of-9x^4+x^3

tangent of

- 9 x^{4} + x^{3}

y = (- 36 a_{0}^{3} + 3 a_{0}^{2}) x + 27 a_{0}^{4} - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 9 a_{0}^{4} + a_{0}^{3})

Find the slope of

y = - 9 x^{4} + x^{3} : \frac{d y}{d x} = - 36 x^{3} + 3 x^{2}

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 = - 36 a_{0}^{3} + 3 a_{0}^{2}

Find the line with slope m=

- 36 a_{0}^{3} + 3 a_{0}^{2}

and passing through

(a_{0}, - 9 a_{0}^{4} + a_{0}^{3}) : y = (- 36 a_{0}^{3} + 3 a_{0}^{2}) x + 27 a_{0}^{4} - 2 a_{0}^{3}

y = (- 36 a_{0}^{3} + 3 a_{0}^{2}) x + 27 a_{0}^{4} - 2 a_{0}^{3}

\int \frac{1}{x ^{3} + 1} d x

\int e^{9 x} cos (2 x) d x

2 t^{2} (\frac{d x}{d t}) = 3 t x + x^{2}

f (x) = 5 sin (x)

\frac{d}{d x} (arccos (3^{x^{2}} - ln (x^{2})))

What is the tangent of-9x^4+x^3 ?
The tangent of-9x^4+x^3 is y=(-36a_{0}^3+3a_{0}^2)x+27a_{0}^4-2a_{0}^3