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

x \to - 10 lim ((+) f (x))

\frac{\partial}{\partial y} (e^{13 x y})

\frac{\partial}{\partial x} (\frac{2 y ^{2}}{x ^{2} + y})

\int p^{3} d p

y^{^{''}} + 9 y = 2 sin (2 t)

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