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 \frac{1}{x ^{2} + 2 x - 24} d x

\frac{\partial}{\partial y} (\frac{( x ^{2} y )}{x ^{2} + y ^{2}})

6 x^{2} + 2 x y + 2 y^{2} = 74, (3, 2)

\int \frac{1}{1 - 36 x ^{2}} d x

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