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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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{5}{x}, at m

n = 0 \sum \infty (\frac{( - 1 ) ^{n}}{2 ^{n}}) x^{2 n}

\frac{d y}{d x} = x (x - 2)

ln (6 x)

(x + y + 4 d x) + (2 x + 2 y + 1 d y) = 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