What is the tangent of y=(-4x)/(x^2+1),(0,0) ?

The tangent of y=(-4x)/(x^2+1),(0,0) is y=-4x

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tangent of y=(-4x)/(x^2+1),(0,0)

tangent of

y = \frac{- 4 x}{x ^{2} + 1}, (0, 0)

y = - 4 x

Solution steps

Find the slope of

y = \frac{- 4 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{4 ( - 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 = - 4

Find the line with slope m=

- 4

and passing through

(0, 0) : y = - 4 x

y = - 4 x

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

\frac{\partial}{\partial x} (x arccos (x))

\int cos (x) \cdot x d x

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

\frac{10}{s ^{2} + 2 \cdot s + 10}

What is the tangent of y=(-4x)/(x^2+1),(0,0) ?
The tangent of y=(-4x)/(x^2+1),(0,0) is y=-4x