What is the tangent of f(x)=-(4x)/(x^2+1),\at x=0 ?

The tangent of f(x)=-(4x)/(x^2+1),\at x=0 is y=-4x

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tangent of f(x)=-(4x)/(x^2+1),\at x=0

tangent of

f (x) = - \frac{4 x}{x ^{2} + 1}, at x = 0

y = - 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = - \frac{4 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = - \frac{4 x}{x ^{2} + 1}

x = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 0) : y = - 4 x

y = - 4 x

\int r^{2} (\frac{r ^{3}}{15} - 8)^{4} d r

\frac{\partial}{\partial x} (\frac{s ^{2} + 4}{t - 3})

\frac{d}{d x} (ln (arctan (2 x^{5})))

\int \frac{1}{6 x ^{5}} d x

3 \cdot 2^{2 x + 1}

What is the tangent of f(x)=-(4x)/(x^2+1),\at x=0 ?
The tangent of f(x)=-(4x)/(x^2+1),\at x=0 is y=-4x