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

\int \frac{1}{2 \cdot x} d x

\frac{1}{x + 3}

e^{- 3 t} cos (2 t)

\int_{0}^{2} 9 x + 37 d x

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

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