What is the tangent of y= 5/(1+x^2),(5,0.19231) ?

The tangent of y= 5/(1+x^2),(5,0.19231) is y=-25/338 x+0.19231+125/338

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= 5/(1+x^2),(5,0.19231)

tangent of

y = \frac{5}{1 + x ^{2}}, (5, 0.19231)

y = - \frac{25}{338} x + 0.19231 + \frac{125}{338}

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{25}{338}

Find the line with slope m=

- \frac{25}{338}

and passing through

(5, 0.19231) : y = - \frac{25}{338} x + 0.19231 + \frac{125}{338}

y = - \frac{25}{338} x + 0.19231 + \frac{125}{338}

\int 9 e^{4 x + e^{4 x}} d x

\int \frac{ln ( x )}{( x - 1 ) ^{2}} d x

y = 12 x, (1, 12)

\int sin (x) d x

(t - 3)^{2}

What is the tangent of y= 5/(1+x^2),(5,0.19231) ?
The tangent of y= 5/(1+x^2),(5,0.19231) is y=-25/338 x+0.19231+125/338