What is the tangent of f(x)=(-4x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(-4x)/(x^2+1),\at x=0 is y=-4x

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tangent of f(x)=(-4x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{- 4 x}{x ^{2} + 1}, at x = 0

y = - 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{- 4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{- 4 x}{x ^{2} + 1}

x = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 0) : y = - 4 x

y = - 4 x

x \to 3 lim (\frac{x ^{2} - 6}{x - 3})

\frac{d ^{2}}{d x ^{2}} (\frac{2}{3 t ^{4}})

ln (1 + 2 \cdot t), 2

\frac{d}{d x} (\frac{4 x - 1}{x ^{2}})

\frac{\partial}{\partial x} (- 6 y^{2} + 12 x y)

What is the tangent of f(x)=(-4x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(-4x)/(x^2+1),\at x=0 is y=-4x