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

The tangent of f(x)= x/(sqrt(9+x^2)),\at x=0 is y= 1/3 x

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

tangent of

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

y = \frac{1}{3} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x}{9 + x ^{2}} : \frac{df ( x )}{d x} = \frac{9}{( 9 + x ^{2} ) 9 + x ^{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}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(0, 0) : y = \frac{1}{3} x

y = \frac{1}{3} x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (5 t)

x^{2} \frac{d y}{d x} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

\int ln (8 x) d x

\int t^{\frac{1}{3}} d t

sin (5 x - 3)

What is the tangent of f(x)= x/(sqrt(9+x^2)),\at x=0 ?
The tangent of f(x)= x/(sqrt(9+x^2)),\at x=0 is y= 1/3 x