What is the tangent of f(x)=(6x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(6x)/(x^2+1),\at x=0 is y=6x

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tangent of f(x)=(6x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{6 x}{x ^{2} + 1}, at x = 0

y = 6 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{6 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{6 x}{x ^{2} + 1}

x = 0 : m = 6

Find the line with slope m=

6

and passing through

(0, 0) : y = 6 x

y = 6 x

\frac{d}{d x} (\frac{π}{x} + ln (2))

\frac{\partial}{\partial x} ((x)^{2} (y)^{4})

\frac{x}{x + 5}

\int (x^{2} - 2) d x

2 x^{2} - 7

What is the tangent of f(x)=(6x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(6x)/(x^2+1),\at x=0 is y=6x