What is the tangent of y=2x^3,(-1,-2) ?

The tangent of y=2x^3,(-1,-2) is y=6x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=2x^3,(-1,-2)

tangent of

y = 2 x^{3}, (- 1, - 2)

y = 6 x + 4

Solution steps

Find the slope of

y = 2 x^{3} : \frac{d y}{d x} = 6 x^{2}

Compute the slope of

y = 2 x^{3}

(- 1, - 2) : m = 6

Find the line with slope m=

6

and passing through

(- 1, - 2) : y = 6 x + 4

y = 6 x + 4

\int tan (y) d y

\int \frac{x ^{6} e ^{x} - 4 x ^{5}}{x ^{6}} d x

x \to π - lim ((x - π) csc (x))

f (x) = \frac{4 x - 7}{28 x ^{5}}

y = cos^{3} (x)