What is the tangent of y= 5/(1+x^2),(4,0.29412) ?

The tangent of y= 5/(1+x^2),(4,0.29412) is y=-40/289 x+0.29412+160/289

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y= 5/(1+x^2),(4,0.29412)

tangent of

y = \frac{5}{1 + x ^{2}}, (4, 0.29412)

y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

Solution steps

Find the slope of

y = \frac{5}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{10 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{40}{289}

Find the line with slope m=

- \frac{40}{289}

and passing through

(4, 0.29412) : y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

What is the tangent of y= 5/(1+x^2),(4,0.29412) ?
The tangent of y= 5/(1+x^2),(4,0.29412) is y=-40/289 x+0.29412+160/289