What is the tangent of f(x)=(-8x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(-8x)/(x^2+1),\at x=0 is y=-8x

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tangent of f(x)=(-8x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 0

y = - 8 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{- 8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{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 = - 8

Find the line with slope m=

- 8

and passing through

(0, 0) : y = - 8 x

y = - 8 x

\int \frac{1}{x} - \frac{3}{x ^{2} + 1} d x

\int (2 sin (x)) d x

\frac{d}{d x} (\frac{3 x ^{3}}{2})

(y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x) + (x e^{x y} cos (2 x) - 3) y^{^{'}} = 0

x \to 0 lim ((1 - 6 x)^{\frac{5}{x}})

What is the tangent of f(x)=(-8x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(-8x)/(x^2+1),\at x=0 is y=-8x