What is the tangent of y=(-8x)/(x^2+1),(1,-4) ?

The tangent of y=(-8x)/(x^2+1),(1,-4) is y=-4

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tangent of y=(-8x)/(x^2+1),(1,-4)

tangent of

y = \frac{- 8 x}{x ^{2} + 1}, (1, - 4)

y = - 4

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}

(1, - 4) : m = 0

Find the line with slope m=

0

and passing through

(1, - 4) : y = - 4

y = - 4

\int csc^{3} (e x) d x

\int ln ∣ x ∣ d x

\int \frac{1}{( x ^{2} - 16 ) ^{2}} d x

y^{^{'}} = y^{2}, y (0) = 2

\frac{d}{d x} ((x^{2} \cdot 5^{x})^{9})

What is the tangent of y=(-8x)/(x^2+1),(1,-4) ?
The tangent of y=(-8x)/(x^2+1),(1,-4) is y=-4