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

The tangent of f(x)=(-8x)/(x^2+1),\at x=1 is y=-4

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

tangent of

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

y = - 4

Solution steps

Find the tangent point:

(1, - 4)

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 = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, - 4) : y = - 4

y = - 4

\frac{\partial}{\partial x} (2 x^{2} y + 4)

(0, 2), (1, 0)

\frac{4 s e ^{- 2 s}}{3 s ^{2} + 12}

\frac{d y}{d t} = y + 2 sin (t) + 1

\int \frac{2}{2 x} d x

What is the tangent of f(x)=(-8x)/(x^2+1),\at x=1 ?
The tangent of f(x)=(-8x)/(x^2+1),\at x=1 is y=-4