What is the tangent of y=x^3+6x-2,(1,5) ?

The tangent of y=x^3+6x-2,(1,5) is y=9x-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^3+6x-2,(1,5)

tangent of

y = x^{3} + 6 x - 2, (1, 5)

y = 9 x - 4

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} + 6 x - 2

(1, 5) : m = 9

Find the line with slope m=

9

and passing through

(1, 5) : y = 9 x - 4

y = 9 x - 4

\frac{d}{d x} (\frac{9}{6 - x})

\int 2 x^{3} + 3 d x

\int_{0}^{1} 0.2 e^{- 0.2 x} d x

\frac{d}{d x} (\frac{x + 1}{x})

What is the tangent of y=x^3+6x-2,(1,5) ?
The tangent of y=x^3+6x-2,(1,5) is y=9x-4