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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 0) : y = - 6 x

y = - 6 x

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