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

The tangent of f(x)= x/(1+x^2),\at x=3 is y=-2/25 x+27/50

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tangent of f(x)= x/(1+x^2),\at x=3

tangent of

f (x) = \frac{x}{1 + x ^{2}}, at x = 3

y = - \frac{2}{25} x + \frac{27}{50}

Solution steps

Find the tangent point:

(3, \frac{3}{10})

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 = 3 : m = - \frac{2}{25}

Find the line with slope m=

- \frac{2}{25}

and passing through

(3, \frac{3}{10}) : y = - \frac{2}{25} x + \frac{27}{50}

y = - \frac{2}{25} x + \frac{27}{50}

5 x^{2} + 3 x^{4}

\frac{( 2 )}{( s - 1 ) ^{2}}

2 cos (5 t)

(s + 2)^{2}

y = 5 + 5 x^{2} - 2 x^{3}

What is the tangent of f(x)= x/(1+x^2),\at x=3 ?
The tangent of f(x)= x/(1+x^2),\at x=3 is y=-2/25 x+27/50