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

The tangent of f(x)= 1/(sqrt(x^2+1)),\at x=2 is y=-2/(5sqrt(5))x+9/(5sqrt(5))

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

tangent of

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

y = - \frac{2}{5 5} x + \frac{9}{5 5}

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = \frac{1}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{x}{( x ^{2} + 1 ) ^{\frac{3}{2}}}

Compute the slope of

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

x = 2 : m = - \frac{2}{5 5}

Find the line with slope m=

- \frac{2}{5 5}

and passing through

(2, \frac{1}{5}) : y = - \frac{2}{5 5} x + \frac{9}{5 5}

y = - \frac{2}{5 5} x + \frac{9}{5 5}

\frac{d}{d x} (ln (5 e^{- 4 x}))

\int \frac{x}{x ^{2} + 4 x + 3} d x

\int \frac{1}{4 - y} d y

x \to 8 + lim (\frac{x - 8}{∣ x - 8 ∣})

y = 3 x^{2} - 2 x, at x = - 1

What is the tangent of f(x)= 1/(sqrt(x^2+1)),\at x=2 ?
The tangent of f(x)= 1/(sqrt(x^2+1)),\at x=2 is y=-2/(5sqrt(5))x+9/(5sqrt(5))