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}}

Compute the slope of

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

x = 0 : 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{\partial}{\partial x} (e^{x} (x^{2} - y^{2}))

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

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

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