What is the tangent of f(x)=(4x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(4x)/(x^2+1),\at x=0 is y=4x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(4x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{4 x}{x ^{2} + 1}, at x = 0

y = 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{4 ( - 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 = 4

Find the line with slope m=

4

and passing through

(0, 0) : y = 4 x

y = 4 x

\int 6 x cos (x) d x

y = x^{2}, y = 3 x

n = 0 \sum \infty 7^{2}

\frac{d y}{d x} = sin (7 x)

x \to \frac{6}{7} lim (x^{4})

What is the tangent of f(x)=(4x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(4x)/(x^2+1),\at x=0 is y=4x