What is the tangent of 5x^3+2x+4,\at x=1 ?

The tangent of 5x^3+2x+4,\at x=1 is y=17x-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 5x^3+2x+4,\at x=1

tangent of

5 x^{3} + 2 x + 4, at x = 1

y = 17 x - 6

Solution steps

Find the tangent point:

(1, 11)

Find the slope of

y = 5 x^{3} + 2 x + 4 : \frac{d y}{d x} = 15 x^{2} + 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 = 17

Find the line with slope m=

17

and passing through

(1, 11) : y = 17 x - 6

y = 17 x - 6

\int \frac{2}{9 - x ^{2}} d x

f (x) = x^{2} sin (4 x^{2})

\frac{\partial}{\partial x} (2 z)

(\frac{4 - 80 y}{y}) (y - 48)

\frac{\partial}{\partial x} (f (x))

What is the tangent of 5x^3+2x+4,\at x=1 ?
The tangent of 5x^3+2x+4,\at x=1 is y=17x-6