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

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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

x \to 5 lim (3 x^{2} - 2 x - 7)

f (x) = 2 x^{2} + 5 x - 3, (1, 4)

\frac{d}{d x} (A x e^{- x} cos (2 x) + B x e^{- x} sin (2 x))

\int_{- 2}^{3} \frac{4}{x ^{4}} d x

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

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