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^{^{'}} = y (1 - 0.001 y)

\int_{0}^{1} π (x^{2})^{2} d x

\frac{d}{d t} (6 t)

f (x) = \frac{x}{x + 6}, at x = 1

f (x) = 3 sec (x), - \frac{π}{6} \leq x \leq \frac{π}{6}

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