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

The tangent of f(x)= 1/(1+x^2),\at x=2 is y=-4/25 x+13/25

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

tangent of

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

y = - \frac{4}{25} x + \frac{13}{25}

Solution steps

Find the tangent point:

(2, \frac{1}{5})

Find the slope of

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

Compute the slope of

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

x = 2 : m = - \frac{4}{25}

Find the line with slope m=

- \frac{4}{25}

and passing through

(2, \frac{1}{5}) : y = - \frac{4}{25} x + \frac{13}{25}

y = - \frac{4}{25} x + \frac{13}{25}

\frac{d x}{d t} = x^{2} + \frac{1}{49}, x (0) = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 6

ln (x + 1)

\frac{d ^{3}}{d x ^{3}} (sin (7 x))

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (1) = 0, y^{^{'}} (1) = 1

What is the tangent of f(x)= 1/(1+x^2),\at x=2 ?
The tangent of f(x)= 1/(1+x^2),\at x=2 is y=-4/25 x+13/25