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

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

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

tangent of

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

y = - 8 x

Solution steps

Find the slope of

f (x) = \frac{- 8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \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

\int_{2}^{5} x d x

\int e^{- \frac{1}{2} z^{2}}

\frac{d x}{d t} = 0.6 - \frac{5 x}{1000 + t}

y = ln (x + 4)

x \to 0 lim (∣ e^{x} - 1 ∣)

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