What is the tangent of y=(-6x)/(x^2+1),(0,0) ?

The tangent of y=(-6x)/(x^2+1),(0,0) is y=-6x

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tangent of y=(-6x)/(x^2+1),(0,0)

tangent of

y = \frac{- 6 x}{x ^{2} + 1}, (0, 0)

y = - 6 x

Solution steps

Find the slope of

y = \frac{- 6 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{- 6 x}{x ^{2} + 1}

(0, 0) : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 0) : y = - 6 x

y = - 6 x

f (x) = 4 x^{2} - x

\int \frac{1}{9 x} d x

\int \frac{x ^{2}}{x ^{2} - 4 x + 20} d x

7 e^{x}, 7 x e^{x^{2}}, 1, 7 e

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

What is the tangent of y=(-6x)/(x^2+1),(0,0) ?
The tangent of y=(-6x)/(x^2+1),(0,0) is y=-6x