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

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

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tangent of y=(-8x)/(x^2+1),(-1,4)

tangent of

y = \frac{- 8 x}{x ^{2} + 1}, (- 1, 4)

y = 4

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{- 8 x}{x ^{2} + 1}

(- 1, 4) : m = 0

Find the line with slope m=

0

and passing through

(- 1, 4) : y = 4

y = 4

x^{2} (240 - \frac{x}{3})

\frac{1}{2} x^{4} - 8

x y \frac{d y}{d x} + 4 x^{2} + y^{2} = 0

\frac{d}{d x} (\frac{5}{2 x ^{2}})

x = y^{2} - 4 y, x = 2 y - y^{2}

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