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

The tangent of y= x/(1+x^2),(2, 2/5) is y=-3/25 x+16/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - \frac{3}{25} x + \frac{16}{25}

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{x}{1 + x ^{2}}

(2, \frac{2}{5}) : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(2, \frac{2}{5}) : y = - \frac{3}{25} x + \frac{16}{25}

y = - \frac{3}{25} x + \frac{16}{25}

f (x) = x + 3

\frac{d y}{d t} = (3 - 2 y) \cdot t^{3}

\frac{\partial}{\partial z} (- 5 z)

y^{^{'}} = 8 y

\frac{d}{d x} (ln (1 + \frac{4}{x}))

What is the tangent of y= x/(1+x^2),(2, 2/5) ?
The tangent of y= x/(1+x^2),(2, 2/5) is y=-3/25 x+16/25