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

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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

\frac{2 ( x + 1 )}{x ^{2} + 1}

r^{2} - 9 x^{2}

N - 0.0002 N^{2} = \frac{d N}{d t}

x \to \infty + lim (x^{10})

\frac{\partial}{\partial x} (\frac{1}{cos ^{2} ( 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