What is the tangent of f(x)= x/(x^2+9),\at x=0 ?

The tangent of f(x)= x/(x^2+9),\at x=0 is y= 1/9 x

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tangent of f(x)= x/(x^2+9),\at x=0

tangent of

f (x) = \frac{x}{x ^{2} + 9}, at x = 0

y = \frac{1}{9} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x}{x ^{2} + 9} : \frac{df ( x )}{d x} = \frac{- x ^{2} + 9}{( x ^{2} + 9 ) ^{2}}

Compute the slope of

f (x) = \frac{x}{x ^{2} + 9}

x = 0 : m = \frac{1}{9}

Find the line with slope m=

\frac{1}{9}

and passing through

(0, 0) : y = \frac{1}{9} x

y = \frac{1}{9} x

y = 3 x^{2} - 4

\frac{d}{d x} (ln (x^{2} - 8))

\int_{5}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

e^{a x} sin (b x)

x \to 3 lim (e^{\frac{2}{3 - x}})

What is the tangent of f(x)= x/(x^2+9),\at x=0 ?
The tangent of f(x)= x/(x^2+9),\at x=0 is y= 1/9 x