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

The tangent of y=(3-2x+4x^3),(-1,1) is y=10x+11

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

y = (3 - 2 x + 4 x^{3}), (- 1, 1)

y = 10 x + 11

Solution steps

Find the slope of

y = (3 - 2 x + 4 x^{3}) : \frac{d y}{d x} = 12 x^{2} - 2

Compute the slope of

y = (3 - 2 x + 4 x^{3})

(- 1, 1) : m = 10

Find the line with slope m=

10

and passing through

(- 1, 1) : y = 10 x + 11

y = 10 x + 11

\frac{d}{d x} (x ln (y) + sin (3 x))

n = 1 \sum \infty (\frac{10}{17})^{n}

\frac{\partial}{\partial x} (cos (x y) y)

\frac{d}{d x} (sin^{c o s (x)} (x))

\int \frac{2 e ^{2 t}}{5 + e ^{2 t}} d t

What is the tangent of y=(3-2x+4x^3),(-1,1) ?
The tangent of y=(3-2x+4x^3),(-1,1) is y=10x+11