What is the tangent of (8x)/(x^2+1),\at x=0 ?

The tangent of (8x)/(x^2+1),\at x=0 is y=8x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of (8x)/(x^2+1),\at x=0

tangent of

\frac{8 x}{x ^{2} + 1}, at x = 0

y = 8 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = \frac{8 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{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 = 8

Find the line with slope m=

8

and passing through

(0, 0) : y = 8 x

y = 8 x

\int (2 x - 4) e^{2 x - 4} d x

\int_{0}^{3} (x 14 - x^{2}) d x

\frac{\partial}{\partial θ} (ρ sin (φ) sin (θ))

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

\frac{\partial}{\partial x} (y (- 3 - x))

What is the tangent of (8x)/(x^2+1),\at x=0 ?
The tangent of (8x)/(x^2+1),\at x=0 is y=8x