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{5 - 4 x ^{3} + 2 x ^{6}}{x ^{6}} d x

\int sin (2 x + 4) d x

\frac{d}{d x} (\frac{sin ( x )}{1 - 2 cos ( x )})

\frac{d y}{d x}, y = 7 x^{2} + 6 y x

\int e^{- θ} cos (8 θ) d θ

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