What is the tangent of y= 2/(1+x^2),(3,0.2) ?

The tangent of y= 2/(1+x^2),(3,0.2) is y=-3/25 x+0.56

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y= 2/(1+x^2),(3,0.2)

tangent of

y = \frac{2}{1 + x ^{2}}, (3, 0.2)

y = - \frac{3}{25} x + 0.56

Solution steps

Find the slope of

y = \frac{2}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{4 x}{( 1 + 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 = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(3, 0.2) : y = - \frac{3}{25} x + 0.56

y = - \frac{3}{25} x + 0.56

What is the tangent of y= 2/(1+x^2),(3,0.2) ?
The tangent of y= 2/(1+x^2),(3,0.2) is y=-3/25 x+0.56