What is the tangent of f(x)= 1/(1+x^2),\at x=-2 ?

The tangent of f(x)= 1/(1+x^2),\at x=-2 is y= 4/25 x+13/25

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tangent of f(x)= 1/(1+x^2),\at x=-2

tangent of

f (x) = \frac{1}{1 + x ^{2}}, at x = - 2

y = \frac{4}{25} x + \frac{13}{25}

Solution steps

Find the tangent point:

(- 2, \frac{1}{5})

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{4}{25}

Find the line with slope m=

\frac{4}{25}

and passing through

(- 2, \frac{1}{5}) : y = \frac{4}{25} x + \frac{13}{25}

y = \frac{4}{25} x + \frac{13}{25}

\int 6 (x - 7)^{2} d x

\int x e^{x} sin (x) d x

\frac{d}{d x} (\frac{3}{7} 5 x^{3})

x \to 0 + lim (7 x ln (x))

x \to 1 lim (e^{7 x^{5} - 7 x})

What is the tangent of f(x)= 1/(1+x^2),\at x=-2 ?
The tangent of f(x)= 1/(1+x^2),\at x=-2 is y= 4/25 x+13/25