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

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

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tangent of y=(-8x)/(x^2+1),(0,0)

tangent of

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

y = - 8 x

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}

(0, 0) : m = - 8

Find the line with slope m=

- 8

and passing through

(0, 0) : y = - 8 x

y = - 8 x

\int \frac{3}{t ^{4}} d t

\int \frac{1}{1 - e ^{x}} d x

\frac{\partial}{\partial x} (x^{4} y^{3} - y^{4})

\frac{d}{d y} (ln (x + 2 y + 3 z))

x \to 5 - lim (\frac{2}{x - 5} + 1)

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