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

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

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

tangent of

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

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{- x ^{2} + 1}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

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

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\int \frac{1}{1 - v} d v

\frac{1}{s ( s + \frac{1}{15} )}

n = 1 \sum \infty \frac{( e ^{n} )}{n !}

\int sin (x - 1) d x

x = 2 y^{2}, x = 4 - y^{2}

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