What is the tangent of f(x)= x/(1+x^2),(2, 2/5) ?

The tangent of f(x)= x/(1+x^2),(2, 2/5) is y=-3/25 x+16/25

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tangent of f(x)= x/(1+x^2),(2, 2/5)

tangent of

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

y = - \frac{3}{25} x + \frac{16}{25}

Solution steps

Find the slope of

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

Compute the slope of

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

(2, \frac{2}{5}) : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(2, \frac{2}{5}) : y = - \frac{3}{25} x + \frac{16}{25}

y = - \frac{3}{25} x + \frac{16}{25}

\frac{1}{6} y^{3} + \frac{1}{2 y}

n = 1 \sum \infty \frac{n}{4 n + 1}

\frac{\partial}{\partial y} (x^{3} - y^{3} + 6 x y)

\frac{d}{d x} (x y^{2} - x^{3} y) = 0

(- 21.2, 22.8), (- 20.2, 7.4)

What is the tangent of f(x)= x/(1+x^2),(2, 2/5) ?
The tangent of f(x)= x/(1+x^2),(2, 2/5) is y=-3/25 x+16/25