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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - 1

Solution steps

Find the slope of

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

Compute the slope of

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

(1, - 1) : m = 0

Find the line with slope m=

0

and passing through

(1, - 1) : y = - 1

y = - 1

\int \frac{x}{64 x ^{2} - 1} d x

\frac{\partial}{\partial x} ((13 - x)^{80})

y (1 + x^{2}) y^{^{'}} = x (1 + y^{2})

\frac{\partial}{\partial x} (\frac{( y - b )}{m})

\int_{0}^{x} \frac{1}{x} d x

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