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{\partial}{\partial x} (\frac{1}{1 + 0 ^{2}})

2 ln (cos (x))

- \frac{y}{2}

\frac{d}{d x} (tan^{2} (sin (x)) + cos (2 x))

f (x) = arctan (\frac{x}{2})

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