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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(2, \frac{2}{5})

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}}

x = 2 : 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}

\int \frac{x}{x - 10} d x

9.86 x^{2}, 4.92 x - x^{2}

\frac{\partial}{\partial t} (t^{3})

5 - \frac{6}{x}

x + 2, 0, 1, 4

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