What is the tangent of y=(-6x)/(x^2+1),(-1,3) ?

The tangent of y=(-6x)/(x^2+1),(-1,3) is y=3

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tangent of y=(-6x)/(x^2+1),(-1,3)

tangent of

y = \frac{- 6 x}{x ^{2} + 1}, (- 1, 3)

y = 3

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}

(- 1, 3) : m = 0

Find the line with slope m=

0

and passing through

(- 1, 3) : y = 3

y = 3

y^{^{''}} + 4 y^{^{'}} - 32 y = 0

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

\frac{d}{d x} (h (g (x) k (x)))

\frac{( s + 1 )}{s ^{2} + s + 1}

\int (x^{3} + 4 x)^{2} (3 x^{2} + 4) d x

What is the tangent of y=(-6x)/(x^2+1),(-1,3) ?
The tangent of y=(-6x)/(x^2+1),(-1,3) is y=3