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

(3 x)^{^{'}}

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

x \to - 2 lim (x^{3} + 4 x^{2} - 5 x + 1)

n = 2 \sum \infty \frac{1}{4 n ^{2} - 1}

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