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

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

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tangent of (-8x)/(x^2+1)(0)

tangent of

\frac{- 8 x}{x ^{2} + 1} (0)

y = 0

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 0)

Find the slope of

y = \frac{- 8 x}{x ^{2} + 1} 0 : \frac{d y}{d x} = 0

Find the line with slope m=

0

and passing through

(a_{0}, 0) : y = 0

y = 0

\int sec^{2} (3 x) d x

\int cos^{2} (4 t) d t

t a y l or x^{- 2}, 5

\frac{d}{d x} (sin (x^{12}))

x \to 0 lim (\frac{- 1}{x})

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