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

The tangent of f(x)= 1/(5+x^2),\at x=2 is y=-4/81 x+17/81

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

tangent of

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

y = - \frac{4}{81} x + \frac{17}{81}

Solution steps

Find the tangent point:

(2, \frac{1}{9})

Find the slope of

f (x) = \frac{1}{5 + x ^{2}} : \frac{df ( x )}{d x} = - \frac{2 x}{( 5 + x ^{2} ) ^{2}}

Compute the slope of

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

x = 2 : m = - \frac{4}{81}

Find the line with slope m=

- \frac{4}{81}

and passing through

(2, \frac{1}{9}) : y = - \frac{4}{81} x + \frac{17}{81}

y = - \frac{4}{81} x + \frac{17}{81}

\int cot^{5} (x) csc (x) d x

\int \frac{x ^{2} - x + 28}{x ^{3} + 7 x} d x

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

n = 1 \sum \infty \frac{1}{1 + 3 ^{n}}

What is the tangent of f(x)= 1/(5+x^2),\at x=2 ?
The tangent of f(x)= 1/(5+x^2),\at x=2 is y=-4/81 x+17/81