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

\int \frac{x ^{2}}{1 + x ^{3}} d x

\int 1 + 9 x^{2} d x

\int \frac{e ^{2 x}}{e ^{2 x} + 7} d x

f (x) = \frac{x}{2 x - 1}

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

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