What is the tangent of f(x)= x/(x^2+81),\at x=0 ?

The tangent of f(x)= x/(x^2+81),\at x=0 is y= 1/81 x

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tangent of f(x)= x/(x^2+81),\at x=0

tangent of

f (x) = \frac{x}{x ^{2} + 81}, at x = 0

y = \frac{1}{81} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x}{x ^{2} + 81} : \frac{df ( x )}{d x} = \frac{- x ^{2} + 81}{( x ^{2} + 81 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{81}

Find the line with slope m=

\frac{1}{81}

and passing through

(0, 0) : y = \frac{1}{81} x

y = \frac{1}{81} x

\int_{- π}^{π} ∣ x ∣ cos (n x) d x

\int_{0}^{1} \frac{x ^{4} + 4}{x} d x

x \to 4 lim (\frac{1}{( 4 - x ) ^{2}})

\frac{d}{d x} (a + b x cos (3 x) + c x sin (3 x))

\frac{\partial}{\partial x} (2 x y - 2 y)

What is the tangent of f(x)= x/(x^2+81),\at x=0 ?
The tangent of f(x)= x/(x^2+81),\at x=0 is y= 1/81 x