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

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

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

tangent of

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

y = 8 x

Solution steps

Find the tangent point:

(0, 0)

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}

x = 0 : m = 8

Find the line with slope m=

8

and passing through

(0, 0) : y = 8 x

y = 8 x

\frac{d}{d x} (x (1 - 2 y))

\frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

f (x) = \frac{2}{x + 3}

\frac{\partial}{\partial y} (x^{y} ln (x))

\int_{0}^{34} x^{2} - 17 d x

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