What is the tangent of y=(8x)/(x^2+1),(0,0) ?

The tangent of y=(8x)/(x^2+1),(0,0) is y=8x

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tangent of y=(8x)/(x^2+1),(0,0)

tangent of

y = \frac{8 x}{x ^{2} + 1}, (0, 0)

y = 8 x

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}

(0, 0) : m = 8

Find the line with slope m=

8

and passing through

(0, 0) : y = 8 x

y = 8 x

(x^{4} + 1) d y + x (1 + 4 y^{2}) d x = 0

n = 0 \sum \infty \frac{n ^{2}}{5 ^{n}}

\int e^{- 2 s} d s

1

\frac{d}{d x} (- \frac{18}{x ^{4}})

What is the tangent of y=(8x)/(x^2+1),(0,0) ?
The tangent of y=(8x)/(x^2+1),(0,0) is y=8x