What is the tangent of y=(-2x)/(x^2+1),(0,0) ?

The tangent of y=(-2x)/(x^2+1),(0,0) is y=-2x

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tangent of y=(-2x)/(x^2+1),(0,0)

tangent of

y = \frac{- 2 x}{x ^{2} + 1}, (0, 0)

y = - 2 x

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}

(0, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

(0, 0) : y = - 2 x

y = - 2 x

f (x) = - \frac{5}{x ^{3}}, y = 0, x = - 4, x = - 1

x \to 0 lim (\frac{x}{csc ^{2} ( x )})

x \to 5 lim (x^{2} - 2 x + 1)

\frac{\partial}{\partial x} (5 e^{yz} cos (x yz))

What is the tangent of y=(-2x)/(x^2+1),(0,0) ?
The tangent of y=(-2x)/(x^2+1),(0,0) is y=-2x