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 θ} (n ln (2 θ))

\int \frac{50 x}{sec ( 5 x ^{2} + 5 )} d x

x \to 0 lim (\frac{( 8 ^{x} - 5 ^{x} )}{x})

\frac{1}{e ^{x^{2}}}

y^{^{'}} = x y^{3} - x y

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