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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, \frac{1}{2}) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

\frac{2}{sin ^{2} ( + 2 )}

y^{^{''}} + y = sec (x) tan (x)

f (x) = x

x \to 0 lim (cos (π x))

\frac{\partial}{\partial x} (6 y)

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