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

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

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

tangent of

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

y = - \frac{1}{2} x + 1

Solution steps

Find the tangent point:

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

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 = 1 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

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

y = - \frac{1}{2} x + 1

\frac{d y}{d x} + y = \frac{1 - e ^{- 2 x}}{e ^{x} - e ^{- x}}

\int x (4 x - 3)^{10} d x

\frac{d}{d x} ((3 x^{2} - 2 x + 3)^{5})

f (x) = ln (2 x + 1)

x^{2} d y + (2 x y - x + 1) d x = 0

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