What is the tangent of f(x)=(2x)/(x^2+1),\at x=0 ?

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

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tangent of f(x)=(2x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{2 x}{x ^{2} + 1}, at x = 0

y = 2 x

Solution steps

Find the tangent point:

(0, 0)

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}

x = 0 : m = 2

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

x \to \infty lim (- 7 x)

\int \frac{1}{x ^{2} + 14 x} d x

\int e^{2 u} d u

\int (2 t - 3 t^{2}) 4 + 36 t^{2} d t

\int x \cdot ln (x^{2}) d x

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