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

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

tangent of

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

y = - 1

Solution steps

Find the slope of

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

Compute the slope of

y = - \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

f (x) = x^{3} - 3 x^{2} + 4 x - 7

\frac{d}{d v} (\frac{u}{u + v})

- x^{3} - 5 x^{2} + 2, at x = - 1

\int cos (x) + 2 d x

\frac{d}{d x} (6 arcsin (x^{2}))

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