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 1 lim (\frac{( 5 - 5 x )}{1 - x})

x \frac{d y}{d x} - y = 0, y (a) = b

y^{^{'}} + 2 x y = x

\int - x arctan (x) d x

\int_{0}^{1} x^{2} + 1 d 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